Topic

Number Theory

Digits, divisibility, factors and primes.

138 problems
Practice
Problem 1 · 2024 AMC 8 Easy
Number Theory last-digit mod-10

What is the ones digit of

222,222 − 22,222 − 2,222 − 222 − 22 − 2 ?
Show hint (soft nudge)
You don't need to do the whole subtraction. What part of the answer is the question actually asking about?
Show hint (sharpest)
Only the ones digits matter — and every number ends in 2. So skip the big subtraction entirely.
Show solution
  1. Only the ones digit matters. The five numbers being subtracted all end in 2, so their ones digits sum to 5 × 2 = 10 — together they take away something ending in 0.
  2. Subtracting a multiple of 10 from 222,222 doesn't touch its ones digit: it stays 2.
B The ones digit is 2.
Another way: keep the intermediate positive (MAA)
  1. Look only at the last two digits so the running total never goes negative: 22 − 2 − 2 − 2 − 2 − 2.
  2. = 22 − 10 = 12. Ones digit: 2.
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Problem 4 · 2024 AMC 8 Medium
Number Theory digit-sum perfect-square work-backward

When Yunji added all the integers from 1 to 9, she mistakenly left out a number. Her incorrect sum turned out to be a square number. What number did Yunji leave out?

Show hint (soft nudge)
Start from the correct total 1+2+…+9. Taking one number out lands you near a special number — which one?
Show hint (sharpest)
Find the correct total of 1 through 9. Leaving out x makes the sum 45 − x. Which x makes that a perfect square?
Show solution
  1. 1 + 2 + 3 + … + 9 = 45.
  2. Leaving out x (from 1 to 9) gives 45 − x, which is between 36 and 44.
  3. The only perfect square in that range is 36 = 62.
  4. 45 − x = 36, so x = 9.
E She left out 9.
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Problem 5 · 2024 AMC 8 Stretch
Number Theory divisibility factor-pairs casework

Aaliyah rolls two standard 6-sided dice. She notices that the product of the two numbers rolled is a multiple of 6. Which of the following integers cannot be the sum of the two numbers?

Show hint (soft nudge)
The product is a multiple of 6 only when the two dice meet a special condition. Test each answer choice against it.
Show hint (sharpest)
A product is a multiple of 6 only if the pair contains a 3 or 6 (factor of 3) and an even number (factor of 2). Just test the answer choices against that.
Show solution
  1. The pair must include a multiple of 3 (a 3 or a 6) and an even number.
  2. Sum 6 comes only from (1,5), (2,4), (3,3) — products 5, 8, 9, none a multiple of 6. So 6 is impossible.
  3. Every other choice has a good pair: 5 = (2,3)→6, 7 = (1,6)→6, 8 = (2,6)→12, 9 = (3,6)→18.
B The sum cannot be 6.
Another way: list every valid pair (MAA)
  1. Pairs whose product is a multiple of 6 (need a multiple of 3 and an even number): (1,6), (2,3), (2,6), (3,6), (4,6), (5,6), (6,6).
  2. Their sums: 7, 5, 8, 9, 10, 11, 12. Among A–E, only 6 is missing.
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Problem 4 · 2023 AMC 8 Stretch
Number Theory primes spiral-pattern
amc8-2023-04
Show hint (soft nudge)
Forget the spiral pattern — what matters is which numbers end up on those four squares.
Show hint (sharpest)
Find the four shaded numbers first (they sit on the diagonal through 7), then test each one for being prime.
Show solution
  1. Continuing the spiral outward, the diagonal through 7 (going up-left and down-right) contains the four shaded numbers 19, 23, 39, 47.
  2. Test each: 19 prime, 23 prime, 47 prime; but 39 = 3 × 13 is composite.
  3. So 3 of the four shaded numbers are prime.
D Three of them are prime.
Another way: use perfect squares as landmarks (MAA)
  1. Without filling the whole grid: on an n×n spiral the number n2 sits in the upper-left (n even) or lower-right (n odd) corner. So 9 is at lower-right of the 3×3 block, 25 at lower-right of 5×5, 49 at lower-right of 7×7; 16 at upper-left of 4×4, 36 at upper-left of 6×6.
  2. Walking outward from those anchors locates the four shaded squares as 19, 23, 39, 47 — with 39 = 3 × 13 the only composite.
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Problem 3 · 2022 AMC 8 Medium
Number Theory factorization factor-triples casework

When three positive integers a, b, and c are multiplied together, their product is 100. Suppose a < b < c. In how many ways can the numbers be chosen?

Show hint (soft nudge)
100 has only a handful of factor groupings. Be systematic so you don't miss any or double-count.
Show hint (sharpest)
List the ways to write 100 as a product of three different whole numbers, smallest to largest. (100 = 22 × 52.)
Show solution
  1. Try the smallest factor as 1: 1×2×50, 1×4×25, 1×5×20 — all have a < b < c.
  2. Now without a 1: 2×5×10 works (2 < 5 < 10).
  3. Others repeat a factor (like 1×10×10 or 2×2×25), so they don't count.
  4. That gives 4 valid choices.
E 4 ways.
Another way: bound the smallest factor first (MAA)
  1. Since a < b < c and abc = 100, we get a3 < 100, so a ≤ 4. And a must be a factor of 100, so a ∈ {1, 2, 4}.
  2. For each: a = 1 gives (1,2,50), (1,4,25), (1,5,20); a = 2 gives (2,5,10); a = 4 gives nothing (would need bc = 25 with 4 < b < c).
  3. Total: 4 ways.
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Problem 5 · 2016 AMC 8 Medium
Number Theory divisibility mod-10

The number N is a two-digit number.

  • When N is divided by 9, the remainder is 1.
  • When N is divided by 10, the remainder is 3.

What is the remainder when N is divided by 11?

Show hint (soft nudge)
Remainder 3 mod 10 means N ends in 3. So N ∈ {13, 23, 33, …, 93}.
Show hint (sharpest)
Which of those leaves remainder 1 when divided by 9? Sum of digits gives the answer fast (sum ≡ N mod 9).
Show solution
  1. Last digit 3, so N ∈ {13, 23, 33, 43, 53, 63, 73, 83, 93}.
  2. Sum-of-digits test for mod 9: only 73 has digit sum 10 ≡ 1 (mod 9). So N = 73.
  3. 73 mod 11: 11 × 6 = 66, remainder 7. Answer: 7.
E Remainder 7.
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Problem 4 · 2014 AMC 8 Easy
Number Theory parity primes

The sum of two prime numbers is 85. What is the product of these two prime numbers?

Show hint
Odd sum from two primes ⇒ one of them must be even. The only even prime is 2.
Show solution
  1. 85 is odd, so one prime is even ⇒ it must be 2.
  2. The other prime: 85 − 2 = 83 (prime).
  3. Product: 2 × 83 = 166.
E 166.
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Problem 8 · 2014 AMC 8 Easy
Number Theory divisibility-by-11

Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $1A2. What is the missing digit A of this 3-digit number?

Show hint (soft nudge)
The total is 11 × (integer), so 1A2 must be divisible by 11.
Show hint (sharpest)
Divisibility rule for 11: alternating sum of digits must be a multiple of 11.
Show solution
  1. 1A2 divisible by 11 ⇒ alternating sum 1 − A + 2 = 3 − A is a multiple of 11.
  2. Single digit A ∈ {0, …, 9}: only A = 3 works (giving 0).
  3. Check: 132 = 11 × 12. ✓
D A = 3.
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Problem 1 · 2013 AMC 8 Easy
Number Theory next-multiple

Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the smallest number of additional cars she must buy in order to be able to arrange all her cars this way?

Show hint
Find the next multiple of 6 above 23.
Show solution
  1. Next multiple of 6 after 23 is 24.
  2. She needs 24 − 23 = 1 more car.
A 1 more car.
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Problem 15 · 2011 AMC 8 Easy
Number Theory exponent-rewrite

How many digits are in the product 45 · 510?

Show hint
Rewrite 45 as 210. Then 210 · 510 = 1010.
Show solution
  1. 45 · 510 = (22)5 · 510 = 210 · 510 = 1010.
  2. 1010 = 1 followed by ten zeros ⇒ 11 digits.
D 11 digits.
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Problem 14 · 2010 AMC 8 Easy
Number Theory prime-factorization

What is the sum of the prime factors of 2010?

Show hint
Factor out small primes first: 2010 / 2 / 3 / 5 = ?
Show solution
  1. 2010 = 2 · 1005 = 2 · 3 · 335 = 2 · 3 · 5 · 67 (67 is prime).
  2. Sum: 2 + 3 + 5 + 67 = 77.
C 77.
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Problem 3 · 2007 AMC 8 Easy
Number Theory prime-factorization

What is the sum of the two smallest prime factors of 250?

Show hint
250 = 2 · 53.
Show solution
  1. 250 = 2 · 53. Two smallest (and only) primes: 2 and 5.
  2. Sum: 2 + 5 = 7.
C 7.
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Problem 10 · 2007 AMC 8 Easy
Number Theory sigma-function

For any positive integer n, define [n] to be the sum of the positive factors of n. For example, [6] = 1 + 2 + 3 + 6 = 12. Find [[11]].

Show hint
[11] first; 11 is prime so [11] = 1 + 11 = 12. Then compute [12].
Show solution
  1. [11] = 1 + 11 = 12 (since 11 is prime).
  2. [12] = 1 + 2 + 3 + 4 + 6 + 12 = 28.
D 28.
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Problem 8 · 2005 AMC 8 Easy
Number Theory parity-rules

Suppose m and n are positive odd integers. Which of the following must also be an odd integer?

Show hint
odd × odd = odd; odd + odd = even; even + odd = odd.
Show solution
  1. (A) odd + odd = even. (B) odd − odd = even. (C) odd + odd = even. (D) (odd + odd)2 = even2 = even.
  2. (E) odd · odd · odd = odd.
E 3mn.
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Problem 18 · 2005 AMC 8 Easy
Number Theory count-multiples

How many three-digit numbers are divisible by 13?

Show hint
Three-digit multiples of 13: smallest is 13 · 8 = 104. Largest is 13 · 76 = 988. Count from 8 to 76.
Show solution
  1. Smallest 3-digit multiple: 13 · 8 = 104. Largest: 13 · 76 = 988.
  2. Multipliers 8 through 76: 76 − 8 + 1 = 69.
C 69.
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Problem 2 · 2003 AMC 8 Easy
Number Theory primes

Which of the following numbers has the smallest prime factor?

Show hint
No prime is smaller than 2 — which of these has 2 as a factor?
Show solution
  1. A prime factor can't be smaller than 2, and only even numbers have 2 as a factor.
  2. 58 is the only even choice, so its smallest prime factor is 2 — smaller than any odd number could manage. Answer 58.
C 58 (its smallest prime factor is 2).
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Problem 4 · 2002 AMC 8 Easy
Number Theory place-value

The year 2002 is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after 2002 that is a palindrome?

Show hint (soft nudge)
Keep the thousands digit at 2, so the next palindrome looks like 2 _ _ 2.
Show hint (sharpest)
For that to mirror, the two middle digits must be equal — make them as small as you can while passing 2002.
Show solution
  1. Don't change the leading 2, so the year stays of the form 2 _ _ 2 with equal middle digits.
  2. The smallest such year after 2002 is 2112, whose digit product is 2 × 1 × 1 × 2 = 4.
B 4.
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Problem 5 · 2002 AMC 8 Easy
Number Theory divisibility

Carlos Montado was born on Saturday, November 9, 2002. On what day of the week will Carlos be 706 days old?

Show hint (soft nudge)
The day of the week repeats every 7 days, so only the remainder of 706 when divided by 7 matters.
Show hint (sharpest)
706 = 700 + 6, and 700 is a multiple of 7.
Show solution
  1. Since 700 is a multiple of 7, after 700 days the weekday is again Saturday.
  2. Six more days past Saturday lands on Friday.
C Friday.
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Problem 4 · 2001 AMC 8 Easy
Number Theory place-value

The digits 1, 2, 3, 4, and 9 are each used once to form the smallest possible even five-digit number. The digit in the tens place is

Show hint (soft nudge)
To make a number small, put the smallest digits on the left — but the units digit has to be even.
Show hint (sharpest)
Push the big digits as far right as the even-ending rule allows.
Show solution
  1. Smallest first three digits are 1, 2, 3. That leaves 4 and 9 for the tens and units.
  2. The units digit must be even, so 4 goes in the units and 9 in the tens: the number is 12394, and the tens digit is 9.
E 9.
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Problem 1 · 1996 AJHSME Easy
Number Theory divisor-counting

How many positive factors of 36 are also multiples of 4?

Show hint (soft nudge)
A factor of 36 that is a multiple of 4 is 4 times a factor of 36 ÷ 4 = 9.
Show hint (sharpest)
List the factors of 9.
Show solution
  1. Such a number is 4 × (a factor of 9), and 9 has factors 1, 3, 9.
  2. That gives 4, 12, 36 — 3 numbers.
B 3.
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Problem 5 · 1986 AJHSME Easy
Number Theory time-arithmetic

A contest began at noon one day and ended 1000 minutes later. At what time did the contest end?

Show hint (soft nudge)
Convert 1000 minutes to hours and minutes.
Show hint (sharpest)
1000 = 16·60 + 40.
Show solution
  1. 1000 minutes = 16 hours 40 minutes.
  2. Noon + 16 h 40 m = 4:40 a.m. (the next day).
D 4:40 a.m.
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Problem 16 · 2026 AMC 8 Hard
Number Theory divisibility-rule

Consider all positive four-digit integers whose digits are all even. What fraction of these integers are divisible by 4?

Show hint (soft nudge)
Divisibility by 4 depends only on the last two digits.
Show hint (sharpest)
With an even tens digit, the tens part is already a multiple of 4, so only the units digit decides it.
Show solution
  1. A number is divisible by 4 exactly when its last two digits are. With an even tens digit, 10·(tens) is already a multiple of 4, so it comes down to the units digit being 0, 4, or 8.
  2. That's 3 of the 5 even units digits, and it holds for every choice of the other digits, so the fraction is 3/5.
D 3/5.
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Problem 18 · 2026 AMC 8 Hard
Number Theory consecutive-sums

In how many ways can 60 be written as the sum of two or more consecutive odd positive integers, arranged in increasing order?

Show hint (soft nudge)
A run of k consecutive odd numbers starting at a sums to k(a + k − 1).
Show hint (sharpest)
Test each k that divides 60 and keep the ones giving a positive odd starting value.
Show solution
  1. k consecutive odd numbers starting at a sum to k(a + k − 1) = 60.
  2. Checking divisors, only k = 2 (29 + 31) and k = 6 (5 + 7 + 9 + 11 + 13 + 15) give a positive odd start.
  3. So there are 2 ways.
B 2.
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Problem 13 · 2025 AMC 8 Medium
Number Theory divisibility spiral-pattern
amc8-2025-13
Show hint (soft nudge)
Don't compute 25 separate remainders — even numbers' remainders mod 7 repeat in a pattern.
Show hint (sharpest)
The cycle is 2, 4, 6, 1, 3, 5, 0 (period 7). 25 numbers = 3 full cycles + 4 extras.
Show solution
  1. The remainders mod 7 of 2, 4, 6, 8, 10, 12, 14, … cycle through 2, 4, 6, 1, 3, 5, 0 with period 7.
  2. 25 even numbers = 3 full cycles (21 numbers, hitting each remainder 3 times) plus 4 extras: 2, 4, 6, 1.
  3. So remainders 1, 2, 4, 6 occur 4 times each; remainders 0, 3, 5 occur 3 times each — the pattern in choice A.
A Histogram (A).
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Problem 16 · 2025 AMC 8 Hard
Number Theory complementary-counting sum-constraint

Five distinct integers from 1 to 10 are chosen, and five distinct integers from 11 to 20 are chosen. No two numbers differ by exactly 10. What is the sum of the ten chosen numbers?

Show hint (soft nudge)
Each chosen low number blocks exactly one high number. How many highs are left to choose from?
Show hint (sharpest)
Five chosen lows block five highs — leaving exactly five unblocked highs, which are forced to be picked. They're the highs matching the unchosen lows + 10.
Show solution
  1. Each chosen low (say x) blocks the high x + 10. With 5 lows chosen, 5 highs are blocked — so the 5 chosen highs are exactly the 5 unblocked ones: those that are 10 more than the 5 unchosen lows.
  2. Sum of chosen highs = (sum of unchosen lows) + 5 × 10. So (chosen lows) + (chosen highs) = (chosen lows) + (unchosen lows) + 50 = (sum of 1 to 10) + 50.
  3. 1 + 2 + … + 10 = 55, so the total is 55 + 50 = 105.
C 105.
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Problem 15 · 2024 AMC 8 Hard
Number Theory factorization casework work-backward

Let the letters F, L, Y, B, U, G represent distinct digits. Suppose FLYFLY is the greatest number that satisfies the equation

8 · FLYFLY = BUGBUG.

What is the value of FLY + BUG?

Show hint (soft nudge)
A six-digit number that repeats a three-digit block (like ABCABC) has a hidden common factor.
Show hint (sharpest)
FLYFLY = 1001 · FLY and BUGBUG = 1001 · BUG, so the equation just says 8 · FLY = BUG.
Show solution
  1. FLYFLY = 1001 · FLY and BUGBUG = 1001 · BUG, so the equation reduces to 8 · FLY = BUG.
  2. BUG is a 3-digit number, so 8 · FLY < 1000 → FLY ≤ 124. So F = 1, and L ≤ 2 (else 8 · FLY ≥ 1040).
  3. Maximize: L = 2 (can't be 1, F took it). Then 8 · 12Y needs all 6 digits distinct. Y = 4: 8 · 124 = 992 (two 9s, fails). Y = 3: 8 · 123 = 984 — digits {1, 2, 3, 9, 8, 4} all distinct ✓.
  4. FLY + BUG = 123 + 984 = 1107.
C 1107.
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Problem 14 · 2023 AMC 8 Hard
Number Theory complementary-counting careful-counting

Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of 5-cent, 10-cent, and 25-cent stamps, with exactly 20 of each type. What is the greatest number of stamps Nicolas can use to make exactly $7.10 in postage?

Show hint (soft nudge)
Maximizing stamps used = minimizing stamps removed from his whole collection. What does the whole collection total?
Show hint (sharpest)
Total of all 60 stamps = $8. He needs to remove $0.90. Minimize the number of stamps that sum to $0.90.
Show solution
  1. Total value of all 60 stamps: 20·($0.05 + $0.10 + $0.25) = 20 · $0.40 = $8.00.
  2. He needs to make $7.10, so he removes $8.00 − $7.10 = $0.90 worth. Maximizing stamps used ≡ minimizing stamps removed.
  3. Minimum stamps summing to $0.90: three 25¢ (75¢) + one 10¢ + one 5¢ = $0.90 in 5 stamps.
  4. Stamps used = 60 − 5 = 55.
E 55 stamps.
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Problem 16 · 2023 AMC 8 Medium
Number Theory divisibility symmetry
amc8-2023-16
Show hint (soft nudge)
20 × 20 = 400 cells, with the letters cycling P, Q, R. How does 400 split among three letters?
Show hint (sharpest)
400 = 3 × 133 + 1, so one letter gets the extra 1. By the table's diagonals, P-count = R-count — so Q takes the extra.
Show solution
  1. The board has 20 × 20 = 400 cells, and the P/Q/R pattern repeats every 3 cells diagonally. 400 = 3 × 133 + 1, so the counts are 133, 133, 134 in some order.
  2. Look at the lower-left 2 × 2 corner of the pattern: Q R / P Q — one P, two Qs, one R. By the rest of the board's symmetry (P and R balance), the corner's extra Q is the surplus.
  3. Counts: 133 Ps, 134 Qs, 133 Rs.
C 133 Ps, 134 Qs, 133 Rs.
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Problem 18 · 2023 AMC 8 Hard
Number Theory divisibility casework

Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump 5 pads to the right or 3 pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located 2023 pads to the right of her starting position?

Show hint (soft nudge)
Order doesn't matter — only the counts of right and left jumps. Set up an equation in R and L.
Show hint (sharpest)
5R − 3L = 2023. Need R ≥ 405 and 5R − 2023 divisible by 3.
Show solution
  1. Order doesn't matter. Let R = right jumps, L = left jumps. Then 5R − 3L = 2023, with R, L ≥ 0.
  2. Minimum R: R ≥ 405 (else 5R < 2023, leaving negative L). And 5R − 2023 must be divisible by 3.
  3. Mod 3: 5R ≡ 2R, and 2023 ≡ 1, so 2R ≡ 1 ⇒ R ≡ 2 (mod 3). The smallest R ≥ 405 satisfying this is 407.
  4. Then L = (5·407 − 2023)/3 = 12/3 = 4. Total: 407 + 4 = 411 jumps.
D 411 jumps.
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Problem 17 · 2022 AMC 8 Medium
Number Theory last-digit mod-10

If n is an even positive integer, the double-factorial notation n!! represents the product of all the even integers from 2 to n. For example: 8!! = 2 × 4 × 6 × 8. What is the units digit of the following sum?

2!! + 4!! + 6!! + … + 2022!!
Show hint (soft nudge)
Once a double-factorial includes 10 as a factor, it ends in 0. Which terms in the sum still affect the ones digit?
Show hint (sharpest)
Only 2!!, 4!!, 6!!, 8!! contribute — everything from 10!! onward ends in 0.
Show solution
  1. n!! for n ≥ 10 contains the factor 10, so its units digit is 0.
  2. Compute only the survivors: 2!! = 2, 4!! = 8, 6!! = 48, 8!! = 384. Their units digits: 2, 8, 8, 4.
  3. Sum of units digits: 2 + 8 + 8 + 4 = 22. Units digit: 2.
B Units digit 2.
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Problem 12 · 2020 AMC 8 Easy
Number Theory factorization

For a positive integer n, the factorial notation n! represents the product of the integers from n to 1. For example: 6! = 6 × 5 × 4 × 3 × 2 × 1. What value of N satisfies the following equation?

5! × 9! = 12 × N!
Show hint (soft nudge)
Compute 5! — can you factor 12 out of it neatly?
Show hint (sharpest)
5! = 120 = 12 × 10. So the equation becomes 12 × 10 × 9! = 12 × N!.
Show solution
  1. 5! = 120 = 12 × 10.
  2. So 5! × 9! = 12 × 10 × 9! = 12 × (10 × 9!) = 12 × 10!.
  3. Therefore N! = 10!, so N = 10.
A N = 10.
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Problem 17 · 2020 AMC 8 Medium
Number Theory factorization divisibility

How many factors of 2020 have more than 3 factors? (As an example, 12 has 6 factors, namely 1, 2, 3, 4, 6, and 12.)

Show hint (soft nudge)
Factor 2020 = 22 × 5 × 101. It has (2+1)(1+1)(1+1) = 12 factors total.
Show hint (sharpest)
Subtract those with at most 3 factors: 1 (one factor), primes 2/5/101 (two each), and squares of primes — here only 4 (three factors).
Show solution
  1. 2020 = 22 · 5 · 101 has (2+1)(1+1)(1+1) = 12 factors.
  2. Factors with ≤ 3 factors: 1 (1 factor); 2, 5, 101 (2 factors each — primes); 4 (3 factors — prime squared). That's 5 of the 12.
  3. Remaining: 12 − 5 = 7 factors with more than 3 factors.
B 7 factors.
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Problem 19 · 2020 AMC 8 Medium
Number Theory divisibility digit-sum

A number is called flippy if its digits alternate between two distinct digits. For example, 2020 and 37373 are flippy, but 3883 and 123123 are not. How many five-digit flippy numbers are divisible by 15?

Show hint (soft nudge)
Divisible by 15 = divisible by 5 AND by 3. The last digit must be 0 or 5; since the number is 5 digits, the first digit can't be 0.
Show hint (sharpest)
Five-digit flippy: pattern ababa. a ≠ 0 and ends in a, so a = 5. Digit sum = 15 + 2b ⇒ b ∈ {0, 3, 6, 9}.
Show solution
  1. 5-digit flippy: ababa with a ≠ b and a ≠ 0.
  2. Div by 5 ⇒ last digit (= a) is 0 or 5. Since a ≠ 0, a = 5.
  3. Number is 5b5b5. Div by 3 ⇒ digit sum 15 + 2b div by 3 ⇒ b div by 3, so b ∈ {0, 3, 6, 9} (and all are ≠ 5).
  4. 4 flippy numbers.
B 4 numbers.
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Problem 13 · 2019 AMC 8 Medium
Number Theory divisibility digit-sum

A palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let N be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of N?

Show hint (soft nudge)
Two-digit palindromes (11, 22, 33, …) are all multiples of 11 — so any sum of three of them is a multiple of 11.
Show hint (sharpest)
Find the smallest 3-digit multiple of 11 that isn't itself a palindrome. Can it be written as a sum of three distinct 2-digit palindromes?
Show solution
  1. Two-digit palindromes (11, 22, …, 99) are all multiples of 11; their sum is too. So N is a multiple of 11.
  2. Smallest 3-digit multiple of 11 that's NOT a palindrome: 110 (palindromes are 121, 131, … — 110 isn't one).
  3. Check: 110 = 11 + 22 + 77 ✓ (three distinct 2-digit palindromes).
  4. N = 110; digit sum = 1 + 1 + 0 = 2.
A 2.
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Problem 14 · 2019 AMC 8 Medium
Number Theory mod-10 divisibility

Isabella has 6 coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every 10 days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the 6 dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?

Show hint (soft nudge)
10 days ≡ 3 days mod 7. So the 6 redemption days are at day-of-week offsets 0, 3, 6, 2, 5, 1 from the starting day.
Show hint (sharpest)
Those 6 offsets cover 6 of 7 days — missing only offset 4. Sunday must be that missing day, so the start = Sunday − 4 days = Wednesday.
Show solution
  1. 10 days advances the day-of-week by 10 mod 7 = 3 days. So the six redemption days have day-of-week offsets {0, 3, 6, 2, 5, 1} mod 7 from the start.
  2. These 6 offsets cover everything except offset 4. Sunday must be that missing offset, so start day is Sunday − 4 days.
  3. Counting back from Sunday: Sat, Fri, Thu, Wed.
C Wednesday.
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Problem 13 · 2018 AMC 8 Hard
Number Theory divisibility casework

Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?

Show hint (soft nudge)
Set 4f + l = 410, with f < l ≤ 100. Get bounds on l first.
Show hint (sharpest)
Since 4f is divisible by 4 and 410 ≡ 2 (mod 4), we need l ≡ 2 (mod 4).
Show solution
  1. Average 82 over 5 tests ⇒ total = 410. Let f = first-four score, l = last. Then 4f + l = 410 with f < l ≤ 100.
  2. l > 82 (else 4f + l would need f ≥ l).
  3. Mod 4: 4f ≡ 0, 410 ≡ 2 ⇒ l ≡ 2 (mod 4). In (82, 100]: l ∈ {86, 90, 94, 98}.
  4. 4 values.
A 4 values.
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Problem 14 · 2018 AMC 8 Medium
Number Theory factorization digit-sum

Let N be the greatest five-digit number whose digits have a product of 120. What is the sum of the digits of N?

Show hint (soft nudge)
Greedy: maximize the leftmost digit first — biggest single digit dividing 120.
Show hint (sharpest)
120 = 8 × 15. 15 = 5 × 3. 3 = 3 × 1. Remaining slots: 1, 1. Number: 85311.
Show solution
  1. Largest digit dividing 120 is 8 (since 9 doesn't divide 120). After taking out 8: leftover product 15.
  2. Largest digit dividing 15 is 5. Leftover: 3. Largest digit dividing 3 is 3. Leftover: 1 → fill remaining digits with 1.
  3. N = 85311. Digit sum: 8 + 5 + 3 + 1 + 1 = 18.
D 18.
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Problem 18 · 2018 AMC 8 Medium
Number Theory factorization prime-test

How many positive factors does 23,232 have?

Show hint (soft nudge)
Find the prime factorization of 23,232, then multiply (exponent + 1) for each prime.
Show hint (sharpest)
23,232 = 26 · 3 · 112.
Show solution
  1. Factor: 23,232 = 2 · 11,616 = 22 · 5,808 = … = 26 · 363 = 26 · 3 · 121 = 26 · 3 · 112.
  2. Number of factors: (6+1)(1+1)(2+1) = 7 · 2 · 3 = 42.
E 42 factors.
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Problem 12 · 2017 AMC 8 Easy
Number Theory divisibility

The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?

Show hint
Same remainder mod 4, 5, AND 6 means n − 1 is divisible by all three. So n − 1 = lcm(4, 5, 6).
Show solution
  1. n − 1 is divisible by 4, 5, and 6, so by lcm(4, 5, 6) = 60.
  2. Smallest n > 1: n = 61, which lies between 60 and 79.
D Between 60 and 79.
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Problem 19 · 2017 AMC 8 Hard
Number Theory factorization divisibility

For any positive integer M, the notation M! denotes the product of the integers 1 through M. What is the largest integer n for which 5n is a factor of the sum

98! + 99! + 100! ?
Show hint (soft nudge)
Factor out 98! from all three terms. The leftover is a clean number.
Show hint (sharpest)
98! + 99! + 100! = 98!(1 + 99 + 99·100) = 98! · 10,000. Count factors of 5 in each piece.
Show solution
  1. 98! + 99! + 100! = 98!(1 + 99 + 99·100) = 98!(100 + 9900) = 98! · 10,000.
  2. 10,000 = 104 = 24 · 54 → contributes 4 factors of 5.
  3. Factors of 5 in 98!: ⌊98/5⌋ + ⌊98/25⌋ = 19 + 3 = 22.
  4. Total: 22 + 4 = 26.
D 26.
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Problem 11 · 2016 AMC 8 Medium
Number Theory place-value careful-counting

Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is 132.

Show hint (soft nudge)
Two-digit + reversed = (10a + b) + (10b + a) = 11(a + b). So a + b = 12.
Show hint (sharpest)
Count digit pairs (a, b) with a + b = 12, a ∈ {1, …, 9}, b ∈ {0, …, 9}.
Show solution
  1. (10a + b) + (10b + a) = 11(a + b) = 132 ⇒ a + b = 12.
  2. Valid pairs with a from 1–9 and b from 0–9: (3,9), (4,8), (5,7), (6,6), (7,5), (8,4), (9,3).
  3. 7 two-digit numbers.
B 7 numbers.
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Problem 15 · 2016 AMC 8 Medium
Number Theory difference-of-squares factorization

What is the largest power of 2 that is a divisor of 134 − 114?

Show hint
Difference of squares twice: a4 − b4 = (a2 + b2)(a2 − b2).
Show solution
  1. 134 − 114 = (132 + 112)(132 − 112) = 290 · 48.
  2. 290 = 2 · 145 (one factor of 2). 48 = 24 · 3 (four factors of 2).
  3. Total: 21+4 = 25 = 32.
C 32.
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Problem 20 · 2016 AMC 8 Medium
Number Theory divisibility factorization

The least common multiple of a and b is 12, and the least common multiple of b and c is 15. What is the least possible value of the least common multiple of a and c?

Show hint (soft nudge)
b divides both lcms — so b divides gcd(12, 15) = 3. Try b = 3.
Show hint (sharpest)
Then minimize a and c: smallest a giving lcm(a, 3) = 12 is 4, and smallest c giving lcm(c, 3) = 15 is 5.
Show solution
  1. b must divide both 12 and 15, so b | gcd(12, 15) = 3.
  2. Take b = 3. Then smallest a with lcm(a, 3) = 12 is a = 4. Smallest c with lcm(c, 3) = 15 is c = 5.
  3. lcm(4, 5) = 20.
A 20.
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Problem 22 · 2015 AMC 8 Hard
Number Theory divisor-count lcm

On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?

Show hint (soft nudge)
Each day's row count is a divisor of n, all different. Twelve days ⇒ n has at least 12 divisors.
Show hint (sharpest)
n is divisible by both 15 and 6, so by lcm(6, 15) = 30. Smallest multiple of 30 with exactly 12 divisors?
Show solution
  1. Each day's number of students per row is a divisor of n, and all 12 are different ⇒ n has exactly 12 divisors (no 13th option exists).
  2. n is a multiple of 15 and 6, hence of lcm(6, 15) = 30 = 2 · 3 · 5. That has only (1+1)(1+1)(1+1) = 8 divisors.
  3. Double one prime exponent to get 12: 22 · 3 · 5 = 60 has (2+1)(1+1)(1+1) = 12 divisors. The other options (32·2·5 = 90, 2·3·52=150) are larger.
  4. Smallest n = 60.
C 60 students.
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Problem 24 · 2015 AMC 8 Hard
Number Theory linear-diophantine mod-3

A baseball league consists of two four-team divisions. Each team plays every other team in its division N games. Each team plays every team in the other division M games with N > 2M and M > 4. Each team plays a 76 game schedule. How many games does a team play within its own division?

Show hint (soft nudge)
A team faces 3 division rivals (N games each) and 4 outside teams (M games each): 3N + 4M = 76.
Show hint (sharpest)
Use the bounds M > 4 and N > 2M; also 4M ≡ 76 (mod 3) forces a residue on M.
Show solution
  1. 3N + 4M = 76.
  2. Mod 3: 4M ≡ 76 ⇒ M ≡ 1 (mod 3). So M ∈ {7, 10, 13, …}.
  3. N > 2M ⇒ 76 = 3N + 4M > 10M ⇒ M < 7.6.
  4. Combining M > 4 and M < 7.6 and M ≡ 1 (mod 3): M = 7. Then N = (76 − 28)/3 = 16.
  5. Games within division: 3N = 3 × 16 = 48.
B 48 games.
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Problem 21 · 2014 AMC 8 Hard
Number Theory divisibility-by-3 mod-arithmetic

The 7-digit numbers 74A52B1 and 326AB4C are each multiples of 3. Which of the following could be the value of C?

Show hint (soft nudge)
Divisibility by 3: sum of digits divisible by 3. Write the digit-sum condition for each number, then subtract.
Show hint (sharpest)
Combine both conditions to find C's residue mod 3.
Show solution
  1. First number: 7+4+5+2+1 = 19, so 19 + A + B ≡ 0 (mod 3) ⇒ A + B ≡ 2 (mod 3).
  2. Second number: 3+2+6+4 = 15, so 15 + A + B + C ≡ 0 (mod 3) ⇒ A + B + C ≡ 0 (mod 3).
  3. Subtract: C ≡ −2 ≡ 1 (mod 3). Choices give C = 1.
A C = 1.
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Problem 22 · 2011 AMC 8 Hard
Number Theory tens-digit-cycle mod-100

What is the tens digit of 72011?

Show hint (soft nudge)
Tens digit depends only on the value mod 100. Compute 7k mod 100 for small k and look for a cycle.
Show hint (sharpest)
71 = 07, 72 = 49, 73 = 343 (43), 74 = 2401 (01). Cycle length 4.
Show solution
  1. Last two digits cycle: 07, 49, 43, 01, repeating with period 4.
  2. 2011 = 4 · 502 + 3 ⇒ 72011 ≡ 73 ≡ 43 (mod 100).
  3. Tens digit = 4.
D Tens digit 4.
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Problem 17 · 2009 AMC 8 Hard
Number Theory exponent-parity-mod

The positive integers x and y are the two smallest positive integers for which the product of 360 and x is a square and the product of 360 and y is a cube. What is the sum of x and y?

Show hint
Prime-factor 360 = 23 · 32 · 5. Make every exponent even (square) or a multiple of 3 (cube).
Show solution
  1. 360 = 23 · 32 · 51.
  2. Square: bump 2's exponent to 4 and 5's to 2 ⇒ multiply by 2 · 5 = 10. So x = 10.
  3. Cube: bump 2's to 3 (already), 3's to 3 (add one 3), 5's to 3 (add two 5s) ⇒ multiply by 3 · 25 = 75. So y = 75.
  4. x + y = 85.
B 85.
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Problem 24 · 2006 AMC 8 Hard
Number Theory factor-1010 cryptarithm

In the multiplication problem below, A, B, C, D are different digits. ABA × CD = CDCD. What is A + B?

Show hint
CDCD = CD × 101. So ABA = 101.
Show solution
  1. CDCD = CD · 101 ⇒ ABA = 101.
  2. So A = 1, B = 0; A + B = 1.
A 1.
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Problem 25 · 2006 AMC 8 Hard
Number Theory parity-of-prime

Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers? (Visible sides: 44, 59, 38.)

Show hint (soft nudge)
Even-visible cards (44, 38) need an odd prime to make an odd sum; the 59 card needs an even prime to make an odd sum. Equal sums ⇒ the 59 card pairs with the only even prime, 2.
Show hint (sharpest)
Common sum = 59 + 2 = 61.
Show solution
  1. Sum behind 59 + 59 = 61 (odd) only if hidden is even ⇒ hidden = 2 (only even prime).
  2. Common sum 61. Behind 44: 61 − 44 = 17 (prime ✓). Behind 38: 61 − 38 = 23 (prime ✓).
  3. Average: (2 + 17 + 23)/3 = 42/3 = 14.
B 14.
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Problem 20 · 2005 AMC 8 Hard
Number Theory modular-meeting

Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise. In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?

Show hint
After k turns, Alice is at +5k and Bob is at −9k (mod 12). They coincide when 5k ≡ −9k (mod 12) ⇒ 14k ≡ 0 (mod 12).
Show solution
  1. 14k ≡ 0 (mod 12) ⇒ 7k ≡ 0 (mod 6) ⇒ k ≡ 0 (mod 6).
  2. Smallest positive: k = 6.
A 6 turns.
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Problem 19 · 2003 AMC 8 Hard
Number Theory divisibility factorization

How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?

Show hint (soft nudge)
A number divisible by all three is divisible by their least common multiple.
Show hint (sharpest)
LCM(15, 20, 25) = 300 — now count its multiples in range.
Show solution
  1. Being divisible by 15, 20, and 25 is the same as being divisible by their LCM.
  2. LCM(15, 20, 25) = 300.
  3. Multiples of 300 between 1000 and 2000: 1200, 1500, 1800 — that's 3.
C 3 integers.
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Problem 14 · 2000 AMC 8 Hard
Number Theory last-digit mod-10

What is the units digit of 1919 + 9999?

Show hint (soft nudge)
Only the units digit of the base matters, and powers of 9 cycle 9, 1, 9, 1, …
Show hint (sharpest)
9 raised to an odd power ends in 9.
Show solution
  1. Powers of 9 end in 9 for odd exponents and 1 for even ones. Both 19 and 99 end in 9, and both exponents are odd, so each power ends in 9.
  2. 9 + 9 = 18, so the units digit is 8.
D 8.
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Problem 10 · 1998 AJHSME Hard
Number Theory casework

Each of the letters W, X, Y, and Z represents a different integer in the set {1, 2, 3, 4}, but not necessarily in that order. If WX − YZ = 1, then the sum of W and Y is

Show hint (soft nudge)
For the difference to be exactly the whole number 1, each fraction should itself be a whole number.
Show hint (sharpest)
Try making one fraction 3/1 and the other 4/2.
Show solution
  1. To get a difference of exactly 1, both fractions should be integers: 3/1 = 3 and 4/2 = 2 work, since 3 − 2 = 1.
  2. Then W = 3 and Y = 4, so W + Y = 7.
E 7.
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Problem 11 · 1997 AJHSME Hard
Number Theory count-divisors
ajhsme-1997-11
Show hint (soft nudge)
The divisor count comes from the prime factorization: add 1 to each exponent and multiply.
Show hint (sharpest)
Work the brackets from the inside out.
Show solution
  1. ⟨11⟩ = 2 (11 is prime) and ⟨20⟩ = (2+1)(1+1) = 6, so the inner value is 2 × 6 = 12.
  2. Then ⟨12⟩ = (2+1)(1+1) = 6.
A 6.
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Problem 15 · 1996 AJHSME Hard
Number Theory units-digit modular-arithmetic

The remainder when the product 1492 · 1776 · 1812 · 1996 is divided by 5 is

Show hint (soft nudge)
Only the units digit of each factor matters for the remainder mod 5.
Show hint (sharpest)
Multiply the units digits and look at that product's units digit.
Show solution
  1. The units digits are 2, 6, 2, 6, and 2 · 6 · 2 · 6 = 144 ends in 4.
  2. A number ending in 4 leaves remainder 4 when divided by 5.
E 4.
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Problem 12 · 1995 AJHSME Hard
Number Theory factor-pairs

A lucky year is one in which at least one date, written as month/day/year, has the property that the month times the day equals the last two digits of the year. For example, 1956 is lucky because 7/8/56 has 7 × 8 = 56. Which of the following is NOT a lucky year?

Show hint (soft nudge)
For each year, try to write its last two digits as month × day with a valid month (1–12) and day.
Show hint (sharpest)
94 has very few factor pairs.
Show solution
  1. 90 = 9 × 10, 91 = 7 × 13, 92 = 4 × 23, 93 = 3 × 31 — all have a month from 1–12 with a real day.
  2. But 94 = 2 × 47 only (besides 1 × 94), and no month 1–12 pairs with a valid day, so 1994 is not lucky.
E 1994.
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Problem 15 · 1995 AJHSME Hard
Number Theory repeating-decimal cyclicity

What is the 100th digit to the right of the decimal point in the decimal form of 4/37?

Show hint (soft nudge)
Write 4/37 as a repeating decimal and find the repeating block.
Show hint (sharpest)
The block has length 3, so reduce 100 by 3.
Show solution
  1. 4/37 = 0.108108… , repeating the block 108 of length 3.
  2. Since 100 = 3·33 + 1, the 100th digit is the 1st of the block: 1.
B 1.
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Problem 22 · 1995 AJHSME Hard
Number Theory factoring

The number 6545 can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?

Show hint (soft nudge)
Factor 6545 into primes first.
Show hint (sharpest)
Group the prime factors into two two-digit numbers.
Show solution
  1. 6545 = 5 · 7 · 11 · 17. Grouping as (5·17)(7·11) = 85 × 77 gives two two-digit numbers.
  2. Their sum is 85 + 77 = 162.
A 162.
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Problem 10 · 1994 AJHSME Hard
Number Theory divisors

For how many positive integer values of N is the expression 36N + 2 an integer?

Show hint (soft nudge)
N + 2 must be a divisor of 36.
Show hint (sharpest)
Since N ≥ 1, only divisors that are at least 3 count.
Show solution
  1. N + 2 must divide 36, and N ≥ 1 means N + 2 ≥ 3.
  2. The divisors of 36 that are ≥ 3 are 3, 4, 6, 9, 12, 18, 36 — 7 of them.
A 7.
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Problem 15 · 1994 AJHSME Hard
Number Theory pattern mod-arithmetic
ajhsme-1994-15
Show hint (soft nudge)
The arrow pattern repeats every 4 steps.
Show hint (sharpest)
Find 425 and 426 modulo 4 to see which arrows apply.
Show solution
  1. The arrows repeat with period 4: from a number ≡ 0 (right), ≡ 1 (up), ≡ 2 (right), ≡ 3 (down).
  2. Since 425 ≡ 1 and 426 ≡ 2 (mod 4), the moves are up then right — choice A.
A Up arrow, then right arrow.
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Problem 15 · 1992 AJHSME Hard
Number Theory periodic-sequence mod

What is the 1992nd letter in the sequence ABCDEDCBAABCDEDCBAABCDEDCBA… ?

Show hint (soft nudge)
The block ABCDEDCBA repeats; count its length.
Show hint (sharpest)
Find 1992 modulo that length.
Show solution
  1. The block ABCDEDCBA has 9 letters and repeats.
  2. 1992 = 9·221 + 3, so the 1992nd letter is the 3rd of the block: C.
C C.
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Problem 9 · 1991 AJHSME Hard
Number Theory inclusion-exclusion

How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?

Show hint (soft nudge)
Count multiples of 3 and of 5 separately.
Show hint (sharpest)
Subtract the multiples of 15, which got counted twice.
Show solution
  1. Multiples of 3: ⌊46/3⌋ = 15; of 5: ⌊46/5⌋ = 9; of 15: ⌊46/15⌋ = 3.
  2. So 15 + 9 − 3 = 21.
B 21.
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Problem 13 · 1991 AJHSME Hard
Number Theory trailing-zeros factor-2-and-5

How many zeros are at the end of the product 25 × 25 × 25 × 25 × 25 × 25 × 25 × 8 × 8 × 8?

Show hint (soft nudge)
Each trailing zero comes from a pair of factors 2 and 5.
Show hint (sharpest)
Count the total 2's and 5's; the smaller count wins.
Show solution
  1. The product is 25⁷ · 8³ = 5¹⁴ · 2⁹, giving fourteen 5's and nine 2's.
  2. Trailing zeros = the smaller count = 9.
C 9.
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Problem 14 · 1989 AJHSME Hard
Number Theory minimize-difference place-value
ajhsme-1989-14
Show hint (soft nudge)
To minimize the difference, make the 3-digit number as small as you can and the 2-digit number as large as you can.
Show hint (sharpest)
Pick the smallest hundreds digit first, then build each number from the digits you have left.
Show solution
  1. Smallest 3-digit using three of {2, 4, 5, 6, 9}: lead with 2, then take the next two smallest ascending → 245.
  2. Largest 2-digit from the remaining {6, 9} is 96. Difference: 245 − 96 = 149.
C 149.
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Problem 16 · 1989 AJHSME Hard
Number Theory parity primes

In how many ways can 47 be written as the sum of two primes?

Show hint (soft nudge)
Odd = odd + even, so one of the two primes must be even.
Show hint (sharpest)
The only even prime is 2 — what does the other prime have to be?
Show solution
  1. 47 is odd, so one prime is even. The only even prime is 2, forcing the other prime to be 47 − 2 = 45.
  2. But 45 = 9 × 5 isn't prime, so there are 0 ways.
A 0.
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Problem 11 · 1988 AJHSME Hard
Number Theory bound-by-perfect-squares

√164 is

Show hint (soft nudge)
Bound 164 between two consecutive perfect squares.
Show hint (sharpest)
12² = 144 and 13² = 169.
Show solution
  1. 12² = 144 < 164 < 169 = 13².
  2. So √164 lies between 12 and 13.
E between 12 and 13.
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Problem 14 · 1988 AJHSME Hard
Number Theory factor-pairs max-sum

◇ and △ are whole numbers and ◇ × △ = 36. The largest possible value of ◇ + △ is

Show hint (soft nudge)
List all factor pairs of 36 and add each pair.
Show hint (sharpest)
Sums are biggest when the factors are most spread out.
Show solution
  1. Factor pairs of 36: (1,36), (2,18), (3,12), (4,9), (6,6). Their sums: 37, 20, 15, 13, 12.
  2. The largest is 1 + 36 = 37.
E 37.
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Problem 20 · 1987 AJHSME Hard
Number Theory counterexample

"If a whole number n is not prime, then the whole number n − 2 is not prime." A value of n which shows this statement to be false is

Show hint (soft nudge)
Look for n that is itself not prime but n − 2 IS prime.
Show hint (sharpest)
Check each composite choice's n − 2.
Show solution
  1. Try n = 9: 9 is not prime, and 9 − 2 = 7 is prime.
  2. That falsifies the claim, so n = 9.
A 9.
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Problem 17 · 1986 AJHSME Hard
Number Theory factor-out parity-of-product

Let o be an odd whole number and let n be any whole number. Which of the following statements about the whole number (o² + no) is always true?

Show hint (soft nudge)
Factor: o² + no = o(o + n).
Show hint (sharpest)
o is odd, so the product is odd iff (o + n) is odd, which means n is even.
Show solution
  1. o(o + n) is odd ⇔ both factors odd. o is already odd, so the product is odd iff o + n is odd, i.e. iff n is even.
  2. So the expression is odd only if n is even — answer E.
E it is odd only if n is even.
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Problem 20 · 1985 AJHSME Hard
Number Theory 31-day-month weekday-frequency

In a certain year, January had exactly four Tuesdays and four Saturdays. On what day did January 1 fall that year?

Show hint (soft nudge)
31 days = 4 weeks + 3 days. Three weekdays appear 5 times, the other four appear 4 times.
Show hint (sharpest)
For Tuesday and Saturday both to land at 4 times, neither can be among the first 3 days of the month.
Show solution
  1. The three weekdays starting on Jan 1 each appear 5 times in a 31-day January. For Tue and Sat both to appear only 4 times, neither can be in those first three weekdays.
  2. Only starting on Wednesday (Wed, Thu, Fri) leaves both Tue and Sat out of that group.
C Wednesday.
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Problem 9 · 2026 AMC 8 Medium
Number Theory simplify-radicals
amc8-2026-09
Show hint (soft nudge)
Work the inner square roots first: √81 = 9 and √16 = 4.
Show hint (sharpest)
Then take the outer square roots of the two products.
Show solution
  1. Inside the radicals, 16√81 = 16 · 9 = 144 and 81√16 = 81 · 4 = 324.
  2. So the value is √144 / √324 = 12 / 18 = 2/3.
B 2/3.
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Problem 6 · 2025 AMC 8 Medium
Number Theory divisibility mod-10

Sekou writes down the numbers 15, 16, 17, 18, 19. After he erases one of his numbers, the sum of the remaining four numbers is a multiple of 4. Which number did he erase?

Show hint (soft nudge)
Adding the five up and testing each removal is slow. What does each number have in common with 4?
Show hint (sharpest)
Look at each number's remainder when divided by 4. The remainder of the whole sum tells you which one to erase.
Show solution
  1. Remainders mod 4: 15→3, 16→0, 17→1, 18→2, 19→3. Their sum is 9, which leaves remainder 1 mod 4.
  2. To make the remaining four sum divisible by 4, erase the one whose remainder is 1 — that's 17.
C 17.
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Problem 9 · 2024 AMC 8 Medium
Number Theory divisibility substitution

All of the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection?

Show hint (soft nudge)
Pick one color to stand for a variable, then write the others in terms of it. The total reveals a hidden factor.
Show hint (sharpest)
Let r = red count. Green = 2r, blue = 4r. Total = 7r — it must be a multiple of 7.
Show solution
  1. Let r be the number of red marbles. Half as many red as green → green = 2r. Twice as many blue as green → blue = 4r.
  2. Total = r + 2r + 4r = 7r — always a multiple of 7.
  3. Among the choices, only 28 = 7 × 4 is a multiple of 7.
E 28 marbles.
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Problem 7 · 2018 AMC 8 Medium
Number Theory divisibility digit-sum

The 5-digit number 2018U is divisible by 9. What is the remainder when this number is divided by 8?

Show hint (soft nudge)
Div by 9 ⇔ digit sum div by 9. Find U.
Show hint (sharpest)
U = 7 (since 2+0+1+8 = 11; 11+U div by 9 ⇒ U = 7). Then 20187 mod 8.
Show solution
  1. Sum of digits 2+0+1+8+U = 11 + U must be a multiple of 9 (with 0 ≤ U ≤ 9). So U = 7.
  2. 20187 ÷ 8 = 2523 remainder 3 (since 2523 × 8 = 20184). Remainder: 3.
B Remainder 3.
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Problem 7 · 2017 AMC 8 Medium
Number Theory factorization divisibility

Let Z be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of Z?

Show hint
Write Z = abcabc as a multiple of abc. The multiplier factors into nice primes.
Show solution
  1. Z = abcabc = abc · 1001.
  2. 1001 = 7 · 11 · 13. So 11 is always a factor of Z.
A 11.
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Problem 8 · 2017 AMC 8 Medium
Number Theory divisibility casework

Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, "My house number has two digits, and exactly three of the following four statements about it are true."

  1. It is prime.
  2. It is even.
  3. It is divisible by 7.
  4. One of its digits is 9.

This information allows Malcolm to determine Isabella's house number. What is its units digit?

Show hint (soft nudge)
Which two statements can't both be true together?
Show hint (sharpest)
A 2-digit number can't be both even and prime (the only even prime is 2). So either (1) or (2) is false — and since 3 of 4 are true, the false one must be (1) "prime".
Show solution
  1. (1) prime and (2) even can't both be true for a 2-digit number (the only even prime is 2). So the false statement is (1), making (2), (3), (4) all true: the number is even, div by 7, and has a 9 as a digit.
  2. Even + divisible by 7 ⇒ divisible by 14. Two-digit multiples of 14: 14, 28, 42, 56, 70, 84, 98. Only 98 has a 9 digit.
  3. Units digit: 8.
D 8.
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Problem 9 · 2017 AMC 8 Medium
Number Theory divisibility casework

All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Marcy could have?

Show hint (soft nudge)
The total must be divisible by both 3 and 4 — so by 12. Try 12 first; if it fails, try 24.
Show hint (sharpest)
12 total: 4 blue + 3 red + 6 green = 13 > 12. Doesn't work. Try 24.
Show solution
  1. Total must be a multiple of 12 (so that 1/3 and 1/4 are integers).
  2. Try 12: blue = 4, red = 3, green = 6 ⇒ total already 13 > 12. Doesn't fit.
  3. Try 24: blue = 8, red = 6, green = 6 ⇒ yellow = 24 − 8 − 6 − 6 = 4.
D 4 yellow marbles.
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Problem 7 · 2016 AMC 8 Medium
Number Theory perfect-square factorization

Which of the following numbers is not a perfect square?

Show hint
Any prime raised to an EVEN power is a perfect square. Any perfect square (like 4) raised to any power is also a perfect square.
Show solution
  1. 12016 = 1 (trivially a square).
  2. 32018 and 52020: even exponents on a prime → perfect squares.
  3. 42019 = (22)2019 = 24038: even exponent overall → perfect square.
  4. 22017: prime with odd exponent → NOT a perfect square.
B 2^2017.
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Problem 9 · 2016 AMC 8 Easy
Number Theory factorization primes

What is the sum of the distinct prime integer divisors of 2016?

Show hint
Factor 2016 into primes.
Show solution
  1. 2016 = 25 · 32 · 7.
  2. Distinct primes: 2, 3, 7. Sum = 2 + 3 + 7 = 12.
B 12.
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Problem 14 · 2015 AMC 8 Medium
Number Theory divisibility algebra-from-pattern

Which of the following integers cannot be written as the sum of four consecutive odd integers?

Show hint (soft nudge)
Write the four consecutive odd integers as n, n+2, n+4, n+6 and add them up. Look for a divisibility pattern in the sum.
Show hint (sharpest)
The sum is 4n + 12 = 4(n + 3). Since n is odd, n + 3 is even, so the sum is divisible by 8.
Show solution
  1. Four consecutive odd integers: n, n+2, n+4, n+6. Sum = 4n + 12 = 4(n + 3).
  2. n is odd, so n + 3 is even ⇒ the sum is a multiple of 8.
  3. Check each choice mod 8: 16, 40, 72, 200 are multiples of 8; 100 is not (100 = 8·12 + 4).
  4. Only 100 fails.
D 100.
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Problem 13 · 2014 AMC 8 Medium
Number Theory parity

If n and m are integers and n2 + m2 is even, which of the following is impossible?

Show hint
Squares preserve parity: k2 is even iff k is even. So n2 + m2 even ⇒ n, m have the same parity.
Show solution
  1. n2 + m2 even ⇒ n2 and m2 have the same parity ⇒ n and m have the same parity.
  2. Same parity ⇒ n + m is always even, never odd.
D n + m cannot be odd.
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Problem 10 · 2013 AMC 8 Medium
Number Theory lcm-gcd prime-factorization

What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?

Show hint
lcm(a, b) / gcd(a, b) = ab / gcd(a, b)2. Or factor and compare exponents.
Show solution
  1. 180 = 22 · 32 · 5. 594 = 2 · 33 · 11.
  2. gcd: 2 · 32 = 18. (Take the minimum exponent of each shared prime.)
  3. lcm: 22 · 33 · 5 · 11 = 4 · 27 · 55 = 5940.
  4. Ratio: 5940 / 18 = 330.
C 330.
Another way: shortcut
  1. lcm(a, b) · gcd(a, b) = a · b, so lcm/gcd = ab/gcd2.
  2. 180 · 594 = 106,920. gcd = 18 ⇒ lcm/gcd = 106,920 / 324 = 330.
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Problem 13 · 2013 AMC 8 Medium
Number Theory place-value-difference divisibility-by-9

When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?

Show hint
If a score's last two digits are 10a + b, reversing gives 10b + a. The difference is 9(a − b): always a multiple of 9.
Show solution
  1. (10a + b) − (10b + a) = 9(a − b).
  2. The sum's error must therefore be a multiple of 9.
  3. Among the choices, only 45 = 9 · 5 qualifies.
A 45.
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Problem 12 · 2012 AMC 8 Medium
Number Theory units-digit-cycle

What is the units digit of 132012?

Show hint
Only the units digit of 13 matters: same as 32012. Powers of 3 cycle: 3, 9, 7, 1, 3, 9, 7, 1, … (period 4).
Show solution
  1. Units digit of 13n equals units digit of 3n.
  2. Cycle length 4: 31→3, 32→9, 33→7, 34→1, then repeats.
  3. 2012 ≡ 0 (mod 4) ⇒ units digit = 1.
A 1.
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Problem 13 · 2012 AMC 8 Medium
Number Theory gcd

Jamar bought some pencils costing more than a penny each at the school bookstore and paid $1.43. Sharona bought some of the same pencils and paid $1.87. How many more pencils did Sharona buy than Jamar?

Show hint
Working in cents: the price (in cents) divides both 143 and 187. Take their gcd; since price > 1 cent, only one option survives.
Show solution
  1. gcd(143, 187) = 11 (since 143 = 11 · 13, 187 = 11 · 17). Price > 1 cent ⇒ price = 11 cents.
  2. Difference Sharona − Jamar = (187 − 143) / 11 = 44 / 11 = 4.
C 4 more pencils.
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Problem 15 · 2012 AMC 8 Medium
Number Theory lcm

The smallest number greater than 2 that leaves a remainder of 2 when divided by 3, 4, 5, or 6 lies between what numbers?

Show hint
If x ≡ 2 (mod each of 3, 4, 5, 6), then x − 2 is a multiple of all of them, so a multiple of lcm(3, 4, 5, 6) = 60.
Show solution
  1. x − 2 divisible by 3, 4, 5, 6 ⇒ divisible by lcm = 60.
  2. Smallest such x > 2 is x = 60 + 2 = 62, which lies between 61 and 65.
D Between 61 and 65.
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Problem 18 · 2012 AMC 8 Medium
Number Theory product-of-distinct-primes

What is the smallest positive integer that is neither prime nor square and that has no prime factor less than 50?

Show hint (soft nudge)
Not prime & no factor < 50 ⇒ product of at least two primes, all ≥ 50. Not a square ⇒ the primes are distinct.
Show hint (sharpest)
Smallest two primes ≥ 50 are 53 and 59.
Show solution
  1. First primes ≥ 50: 53, 59, 61, …
  2. Smallest product of two distinct such primes: 53 · 59 = 3127.
A 3127.
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Problem 17 · 2011 AMC 8 Medium
Number Theory prime-factorization

Let w, x, y, and z be whole numbers. If 2w · 3x · 5y · 7z = 588, then what does 2w + 3x + 5y + 7z equal?

Show hint
Factor 588 into primes: 588 = 4 · 147 = 4 · 3 · 49.
Show solution
  1. 588 = 22 · 31 · 50 · 72 ⇒ w = 2, x = 1, y = 0, z = 2.
  2. 2w + 3x + 5y + 7z = 4 + 3 + 0 + 14 = 21.
A 21.
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Problem 24 · 2011 AMC 8 Medium
Number Theory parity primes

In how many ways can 10001 be written as the sum of two primes?

Show hint
Odd sum ⇒ one prime must be even ⇒ that prime is 2. Then check 10001 − 2 = 9999.
Show solution
  1. 10001 is odd; primes summing to odd require one to be even, so it must be 2.
  2. Then the other is 10001 − 2 = 9999 = 3 · 3333 (digit sum 36 divisible by 3). Not prime.
  3. So 0 representations.
A 0 ways.
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Problem 11 · 2009 AMC 8 Medium
Number Theory gcd

The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of $1.43. Some of the 30 sixth graders each bought a pencil, and they paid a total of $1.95. How many more sixth graders than seventh graders bought a pencil?

Show hint
Working in cents, the price divides both 143 and 195. Compute gcd(143, 195).
Show solution
  1. 143 = 11 · 13; 195 = 3 · 5 · 13. So price | 13 and price > 1 ⇒ price = 13¢.
  2. Seventh graders: 143/13 = 11. Sixth graders: 195/13 = 15.
  3. Difference: 15 − 11 = 4.
D 4 more.
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Problem 15 · 2008 AMC 8 Medium
Number Theory divisibility average-as-integer

In Theresa's first 8 basketball games, she scored 7, 4, 3, 6, 8, 3, 1 and 5 points. In her ninth game, she scored fewer than 10 points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than 10 points and her points-per-game average for the 10 games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?

Show hint (soft nudge)
Sum of first 8: 37. After game 9 (score < 10), total is between 37 and 47; must be a multiple of 9 (mean integer).
Show hint (sharpest)
Then after game 10 the total is < 56 and a multiple of 10.
Show solution
  1. Sum after 8: 37.
  2. After 9: total in [38, 47], divisible by 9 ⇒ 45. Game 9 = 8.
  3. After 10: total in [46, 55], divisible by 10 ⇒ 50. Game 10 = 5.
  4. Product: 8 · 5 = 40.
B 40.
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Problem 22 · 2008 AMC 8 Medium
Number Theory range-of-integers

For how many positive integer values of n are both n3 and 3n three-digit whole numbers?

Show hint
Let x = n/3 (so n = 3x). Then 3n = 9x. Both x and 9x are 3-digit.
Show solution
  1. 100 ≤ x ≤ 999 and 100 ≤ 9x ≤ 999.
  2. The second is the binding constraint: x ≤ 111.
  3. Combined with x ≥ 100: x ∈ {100, 101, …, 111} — 12 values.
A 12.
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Problem 18 · 2007 AMC 8 Medium
Number Theory last-digits-only

The product of the two 99-digit numbers 303,030,303,…,030,303 and 505,050,505,…,050,505 has thousands digit A and units digit B. What is the sum of A and B?

Show hint
The last 4 digits of each factor are 0303 and 0505. Multiply those mod 10000.
Show solution
  1. 303 · 505 = 153015.
  2. Last 4 digits: 3015 ⇒ thousands digit A = 3, units digit B = 5.
  3. A + B = 8.
D 8.
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Problem 11 · 2006 AMC 8 Medium
Number Theory digit-sum casework

How many two-digit numbers have digits whose sum is a perfect square?

Show hint (soft nudge)
Digit sums of two-digit numbers range 1–18. Perfect squares in range: 1, 4, 9, 16.
Show hint (sharpest)
Count two-digit numbers with each of these digit sums.
Show solution
  1. Sum = 1: {10} ⇒ 1.
  2. Sum = 4: {13, 22, 31, 40} ⇒ 4.
  3. Sum = 9: {18, 27, 36, 45, 54, 63, 72, 81, 90} ⇒ 9.
  4. Sum = 16: {79, 88, 97} ⇒ 3.
  5. Total: 1 + 4 + 9 + 3 = 17.
C 17.
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Problem 23 · 2006 AMC 8 Medium
Number Theory chinese-remainder-by-listing

A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people?

Show hint
List numbers ≡ 4 (mod 6) and find the first one that's also ≡ 3 (mod 5).
Show solution
  1. ≡ 4 (mod 6): 4, 10, 16, 22, 28, …
  2. ≡ 3 (mod 5): 3, 8, 13, 18, 23, 28, …
  3. First common: 28. 28 / 7 = 4 remainder 0.
A 0.
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Problem 19 · 2004 AMC 8 Medium
Number Theory lcm

A whole number larger than 2 leaves a remainder of 2 when divided by each of the numbers 3, 4, 5, and 6. The smallest such number lies between which two numbers?

Show hint
x − 2 is divisible by lcm(3, 4, 5, 6) = 60.
Show solution
  1. Smallest x > 2 with x − 2 = 60 ⇒ x = 62.
  2. 62 lies between 60 and 79.
B Between 60 and 79.
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Problem 9 · 2000 AMC 8 Medium
Number Theory powers careful-counting
amc8-2000-09
Show hint (soft nudge)
List all the three-digit powers of 5, and of 2.
Show hint (sharpest)
The shared crossing digit forces which power of 2 fits.
Show solution
  1. Three-digit powers of 5 are only 125 and 625; three-digit powers of 2 are 128, 256, and 512.
  2. The crossing forces the power of 2 to be 256, so the outlined square holds its units digit, 6.
D 6.
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Problem 11 · 2000 AMC 8 Medium
Number Theory divisibility casework

The number 64 has the property that it is divisible by its units digit. How many whole numbers between 10 and 50 have this property?

Show hint (soft nudge)
Group the numbers by their units digit and test divisibility.
Show hint (sharpest)
Units digit 1, 2, or 5 always works; 0 never does (no division by 0).
Show solution
  1. Units digit 1, 2, and 5 each give 4 working numbers (12 total). Then 33 (digit 3), 24 and 44 (digit 4), 36 (digit 6), and 48 (digit 8) work; digits 0, 7, 9 give none.
  2. Total = 12 + 1 + 2 + 1 + 1 = 17.
C 17.
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Problem 5 · 1997 AJHSME Medium
Number Theory digit-sum listing

There are many two-digit multiples of 7, but only two of them have a digit sum of 10. The sum of these two multiples of 7 is

Show hint (soft nudge)
List the two-digit multiples of 7 and check each digit sum.
Show hint (sharpest)
Find the two whose digits add to 10.
Show solution
  1. Among 14, 21, 28, …, 98, the ones with digit sum 10 are 28 and 91.
  2. Their sum is 28 + 91 = 119.
A 119.
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Problem 5 · 1996 AJHSME Medium
Number Theory sign-analysis
ajhsme-1996-05
Show hint (soft nudge)
Read each letter's sign and rough size from the number line.
Show hint (sharpest)
A difference is negative when you subtract a larger number from a smaller one.
Show solution
  1. From the line, P ≈ −2.7 and Q ≈ −1, while R, S, T are positive.
  2. P − Q ≈ −2.7 − (−1) = −1.7 is negative; the other four expressions all come out positive.
  3. So P − Q is the negative one.
A P − Q.
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Problem 7 · 1996 AJHSME Medium
Number Theory powers exponential

Brent has goldfish that quadruple (become four times as many) every month, and Gretel has goldfish that double every month. If Brent has 4 goldfish at the same time that Gretel has 128 goldfish, in how many months from that time will they have the same number of goldfish?

Show hint (soft nudge)
Write both counts as powers of 2 and match the exponents.
Show hint (sharpest)
Brent: 4 · 4^m; Gretel: 128 · 2^m.
Show solution
  1. After m months Brent has 4 · 4^m = 2^(2m+2) and Gretel has 128 · 2^m = 2^(m+7).
  2. Setting 2m + 2 = m + 7 gives m = 5.
B 5 months.
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Problem 8 · 1996 AJHSME Medium
Number Theory number-line extremal

Points A and B are 10 units apart. Points B and C are 4 units apart. Points C and D are 3 units apart. If A and D are as close as possible, then the number of units between them is

Show hint (soft nudge)
Walk from A: out 10, then come back as much as possible.
Show hint (sharpest)
The best you can backtrack is 4 + 3 = 7.
Show solution
  1. Place B 10 from A, then step C back 4 and D back another 3: A is at 0, D ends at 10 − 4 − 3 = 3.
  2. You can't reach 0 (10 ≠ 4 + 3), so the closest is 3 units.
B 3 units.
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Problem 6 · 1994 AJHSME Medium
Number Theory divisibility

The units digit (one's digit) of the product of any six consecutive positive whole numbers is

Show hint (soft nudge)
Among six consecutive numbers, what's guaranteed to appear?
Show hint (sharpest)
A multiple of 5 and an even number together make a multiple of 10.
Show solution
  1. Any six consecutive numbers include a multiple of 5 and at least one even number.
  2. Their product is therefore a multiple of 10, so its units digit is 0.
A 0.
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Problem 3 · 1993 AJHSME Medium
Number Theory prime-factorization

Which of the following numbers has the largest prime factor?

Show hint (soft nudge)
Factor each number into primes.
Show hint (sharpest)
Then compare the biggest prime in each.
Show solution
  1. 39 = 3·13, 51 = 3·17, 77 = 7·11, 91 = 7·13, 121 = 11².
  2. The largest prime factor among these is 17, from 51.
B 51.
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Problem 7 · 1993 AJHSME Medium
Number Theory exponent-rules

33 + 33 + 33 =

Show hint (soft nudge)
Three copies of the same thing is 3 times it.
Show hint (sharpest)
3 × 3³ = 3¹ × 3³.
Show solution
  1. 3³ + 3³ + 3³ = 3 × 3³.
  2. Adding exponents, 3 × 3³ = 3⁴.
A 3⁴.
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Problem 1 · 1990 AJHSME Medium
Number Theory place-value minimize
ajhsme-1990-01
Show hint (soft nudge)
The hundreds digits matter most, so put the two smallest there.
Show hint (sharpest)
Then the next-smallest in the tens, and the largest in the units.
Show solution
  1. Put 4 and 5 in the hundreds (900), 6 and 7 in the tens (130), and 8 and 9 in the units (17).
  2. The smallest sum is 900 + 130 + 17 = 1047.
C 1047.
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Problem 4 · 1990 AJHSME Medium
Number Theory units-digit-of-squares

Which of the following could not be the units digit (one's digit) of the square of a whole number?

Show hint (soft nudge)
Square the digits 0 through 9 and look at the last digit.
Show hint (sharpest)
Some digits never show up as a final digit.
Show solution
  1. Squares end only in 0, 1, 4, 5, 6, or 9.
  2. 8 is not on that list, so a square can never end in 8.
E 8.
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Problem 10 · 1988 AJHSME Medium
Number Theory mod-7

Chris's birthday is on a Thursday this year. What day of the week will it be 60 days after her birthday?

Show hint (soft nudge)
Days of the week repeat every 7 days, so reduce 60 modulo 7.
Show hint (sharpest)
60 = 8·7 + 4.
Show solution
  1. 60 days = 8 full weeks plus 4 extra days.
  2. 4 days after Thursday → Friday, Saturday, Sunday, Monday.
A Monday.
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Problem 8 · 1987 AJHSME Medium
Number Theory bound-the-sum
ajhsme-1987-08
Show hint (soft nudge)
Check the smallest and largest possible sums to see whether the digit count can change.
Show hint (sharpest)
Smallest: A = B = 1. Largest: A = B = 9.
Show solution
  1. Smallest: 9876 + 132 + 11 = 10019. Largest: 9876 + 932 + 91 = 10899.
  2. Both have 5 digits, so the sum always does.
B 5.
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Problem 9 · 1987 AJHSME Medium
Number Theory lcm-from-prime-factors

When finding the sum 1⁄2 + 1⁄3 + 1⁄4 + 1⁄5 + 1⁄6 + 1⁄7, the least common denominator used is

Show hint (soft nudge)
Take the highest power of each prime in 2, 3, 4, 5, 6, 7.
Show hint (sharpest)
4 contributes 2², the rest contribute 3, 5, and 7 once.
Show solution
  1. Primes appearing: 2² (from 4), 3, 5, 7.
  2. LCM = 4 × 3 × 5 × 7 = 420.
C 420.
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Problem 7 · 1986 AJHSME Medium
Number Theory bound-square-roots

How many whole numbers are between √8 and √80?

Show hint (soft nudge)
√8 sits between 2 and 3; √80 sits between 8 and 9.
Show hint (sharpest)
Count the integers strictly between.
Show solution
  1. 4 < 8 < 9, so 2 < √8 < 3. 64 < 80 < 81, so 8 < √80 < 9.
  2. Whole numbers strictly between √8 and √80 are 3, 4, 5, 6, 7, 8 — that's 6.
B 6.
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Problem 8 · 1986 AJHSME Medium
Number Theory ones-digit guess-and-check
ajhsme-1986-08
Show hint (soft nudge)
The ones digit of the product is 2 · B, which must end in 6.
Show hint (sharpest)
Then check that B2 × 7B actually equals 6396.
Show solution
  1. Product ends in 6, and 2 × B ends in 6 only for B = 3 or B = 8.
  2. Test: 82 × 78 = 6396. ✓ So B = 8.
E 8.
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Problem 24 · 2026 AMC 8 Stretch
Number Theory legendre-formula

The notation n! is the product of the first n positive integers. Define the superfactorial of n to be the product of the factorials 1! · 2! · 3! · … · n! (so the superfactorial of 3 is 1! · 2! · 3! = 12). How many factors of 7 appear in the prime factorization of the superfactorial of 51?

Show hint (soft nudge)
The 7-count of the superfactorial is the sum of the 7-counts of 1! through 51!.
Show hint (sharpest)
Each k! holds ⌊k/7⌋ + ⌊k/49⌋ sevens; group the values of k by that count.
Show solution
  1. The number of 7s is the sum of v₇(k!) for k = 1 to 51, where v₇(k!) = ⌊k/7⌋ + ⌊k/49⌋.
  2. Seven values of k each contribute 1, 2, 3, 4, 5, and 6, totaling 7(1+2+3+4+5+6) = 147; and k = 49, 50, 51 contribute 8 each, adding 24.
  3. The total is 147 + 24 = 171.
E 171.
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Problem 22 · 2025 AMC 8 Hard
Number Theory factorization factor-pairs
amc8-2025-22
Show hint (soft nudge)
Imagine adding a phantom coat after the last one. Now the row is a clean repeating pattern.
Show hint (sharpest)
After the phantom, you have 36 hooks split into d identical blocks of b hooks each, with b ≥ 2 and d ≥ 2. How many (b, d) factorizations of 36 are there?
Show solution
  1. Add a phantom coat after the last real coat. Now 36 hooks form a perfectly repeating pattern: each block is (some empty hooks) + (one coat), of length b ≥ 2.
  2. If there are d blocks, then bd = 36, and the real coat count = d − 1 (we added one).
  3. Both b ≥ 2 (each block has ≥ 1 empty + 1 coat) and d ≥ 2 (at least 1 original coat + the phantom).
  4. 36 = 22 × 32 has (2+1)(2+1) = 9 divisors. Removing the two ordered factorizations with a 1 (1×36 and 36×1) leaves 9 − 2 = 7.
D 7 different coat counts.
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Problem 23 · 2025 AMC 8 Hard
Number Theory primes difference-of-squares prime-test

How many four-digit numbers have all three of the following properties?

  1. The tens digit and ones digit are both 9.
  2. The number is 1 less than a perfect square.
  3. The number is the product of exactly two prime numbers.
Show hint (soft nudge)
The number ends in 99, so the perfect square just above it ends in 00. What does that say about its square root?
Show hint (sharpest)
Use a2 − 1 = (a − 1)(a + 1). For the number to be a product of exactly two primes, both factors must be prime — twin primes.
Show solution
  1. A 4-digit number ending in 99 plus 1 ends in 00. So that perfect square = (10k)2, and our number is (10k)2 − 1 = (10k − 1)(10k + 1).
  2. Four-digit range gives 10k ∈ {40, 50, 60, 70, 80, 90, 100}.
  3. We need both 10k − 1 and 10k + 1 prime. Check: 39×41 (39 = 3×13, no), 49×51 (49 = 72, no), 59×61 (both prime ✓), 69×71 (69 = 3×23, no), 79×81 (81 = 34, no), 89×91 (91 = 7×13, no), 99×101 (99 = 9×11, no).
  4. Only 59 × 61 = 3599 works. Exactly 1.
B Exactly 1.
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Problem 23 · 2024 AMC 8 Hard
Number Theory factorization grid grid-counting
amc8-2024-23
Show hint (soft nudge)
The example (0,4) to (2,0) crosses 4 cells. Notice: 2 + 4 − gcd(2,4) = 4. That's a general formula.
Show hint (sharpest)
Apply the formula to (3000, 5000) horizontal/vertical offsets: 3000 + 5000 − gcd(3000, 5000).
Show solution
  1. A line segment whose horizontal offset is a and vertical offset is b crosses a + b − gcd(a, b) grid cells. (You'd cross a + b cells if the line never hit a grid corner; each lattice-point crossing collapses two cell-entries into one, saving 1 per shared factor.)
  2. From (2000, 3000) to (5000, 8000): horizontal offset 3000, vertical offset 5000.
  3. gcd(3000, 5000) = 1000. Cells = 3000 + 5000 − 1000 = 7000.
C 7000 cells.
Another way: scale down (MAA)
  1. The slope from (2000, 3000) to (5000, 8000) is 5/3. The segment is equivalent to 1000 copies of a primitive (0,0)→(3,5) piece, since gcd(3000, 5000) = 1000.
  2. Each primitive (3,5) segment crosses 7 cells. Total: 7 × 1000 = 7000.
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Problem 22 · 2023 AMC 8 Hard
Number Theory factorization substitution

In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term in the sequence is 4000. What is the first term?

Show hint (soft nudge)
Write the first six terms as a, b, and powers of a and b. The 6th term will be a3b5.
Show hint (sharpest)
Factor 4000: 4000 = 53 × 25. Match exponents.
Show solution
  1. Let the first two terms be a, b. Then the next four are ab, ab2, a2b3, a3b5. (Each term sums the exponents of the previous two.)
  2. a3b5 = 4000 = 53 × 25. So a = 5, b = 2.
  3. First term: 5.
D 5.
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Problem 22 · 2020 AMC 8 Hard
Number Theory work-backward casework

When a positive integer N is fed into a machine, the output is calculated by the rule: if N is even, output N/2; if N is odd, output 3N + 1. Example: 7 → 22 → 11 → 34 → 17 → 52 → 26. When the same 6-step process is applied to a different starting N, the final output is 1. What is the sum of all such integers N?

Show hint (soft nudge)
Work backward from 1. The predecessors of any k are: 2k (always), and (k − 1)/3 (only if that's an odd integer).
Show hint (sharpest)
Build the inverse-tree six levels up from 1. The level-6 leaves are the valid starting values.
Show solution
  1. Forward: even → halve, odd → 3n+1. Inverting from a value k: predecessors are 2k (always), and (k−1)/3 (only if that's an odd integer).
  2. Walk backward from 1: 1 → {2} → {4} → {8, 1} → {16, 2} → {32, 5, 4} → {64, 10, 8, 1}.
  3. Sum: 64 + 10 + 8 + 1 = 83.
E Sum is 83.
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Problem 21 · 2018 AMC 8 Hard
Number Theory divisibility casework

How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?

Show hint (soft nudge)
Each remainder is exactly 4 less than the divisor (2 = 6−4, 5 = 9−4, 7 = 11−4). So x ≡ −4 mod each of them.
Show hint (sharpest)
x + 4 is divisible by lcm(6, 9, 11) = 198. Count multiples of 198 with x + 4 in [104, 1003].
Show solution
  1. 2 = 6 − 4, 5 = 9 − 4, 7 = 11 − 4 ⇒ x ≡ −4 mod 6, 9, 11 simultaneously.
  2. So x + 4 is a multiple of lcm(6, 9, 11) = 2 · 32 · 11 = 198.
  3. 100 ≤ x ≤ 999 ⇒ 104 ≤ x + 4 ≤ 1003. Multiples of 198 in that range: 198, 396, 594, 792, 990. 5 values.
E 5 integers.
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Problem 25 · 2018 AMC 8 Hard
Number Theory perfect-square estimate-and-pick

How many perfect cubes lie between 28 + 1 and 218 + 1, inclusive?

Show hint (soft nudge)
218 = (26)3 = 643. So cubes ≤ 218 + 1 means base ≤ 64.
Show hint (sharpest)
28 = 256, 28 + 1 = 257. The next cube above 257 is 73 = 343 (since 63 = 216).
Show solution
  1. Upper end: 218 = (26)3 = 643, so 643 ≤ 218 + 1 ✓. Cubes at base ≥ 65 exceed the range.
  2. Lower end: 63 = 216 < 257 = 28 + 1, but 73 = 343 ≥ 257 ✓. Smallest valid base: 7.
  3. Count integers from 7 to 64 inclusive: 64 − 7 + 1 = 58.
E 58 cubes.
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Problem 23 · 2017 AMC 8 Hard
Number Theory factorization arithmetic-sequence

Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by 5 minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?

Show hint (soft nudge)
Each day's min-per-mile must divide 60 (so that 60 min yields an integer mile count). Find four divisors of 60 in arithmetic progression with common difference 5.
Show hint (sharpest)
Divisors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. The only arithmetic-progression-5 of length 4 is 5, 10, 15, 20.
Show solution
  1. 60 minutes total per day; for integer miles, the min-per-mile divides 60.
  2. Four divisors of 60 forming an AP with common difference 5: 5, 10, 15, 20.
  3. Miles each day: 60/5 + 60/10 + 60/15 + 60/20 = 12 + 6 + 4 + 3 = 25.
C 25 miles.
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Problem 24 · 2017 AMC 8 Hard
Number Theory complementary-counting divisibility

Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?

Show hint (soft nudge)
The pattern repeats every lcm(3, 4, 5) = 60 days. Count call-days in 60, then scale to 365 (with the leftover 5 days handled).
Show hint (sharpest)
Inclusion–exclusion in a 60-day block: 20 + 15 + 12 − 5 − 4 − 3 + 1 = 36 call days. So 60 − 36 = 24 no-call days.
Show solution
  1. lcm(3, 4, 5) = 60, so the call pattern repeats every 60 days. In one 60-day block:
  2. Calls by inclusion-exclusion: 20 (every 3) + 15 (every 4) + 12 (every 5) − 5 (every 12) − 4 (every 15) − 3 (every 20) + 1 (every 60) = 36 call days. So 60 − 36 = 24 no-call days per block.
  3. 365 days = 6 full 60-day blocks + 5 extra days. No-call days from blocks: 6 × 24 = 144.
  4. In the last 5 days (days 361–365 ≡ days 1–5 of a new cycle): day 1 no call, day 2 no call, day 3 call, day 4 call, day 5 call. So 2 more no-call days.
  5. Total: 144 + 2 = 146.
D 146 days.
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Problem 24 · 2016 AMC 8 Hard
Number Theory divisibility casework

The digits 1, 2, 3, 4, and 5 are each used once to write a five-digit number PQRST. The three-digit number PQR is divisible by 4, the three-digit number QRS is divisible by 5, and the three-digit number RST is divisible by 3. What is P?

Show hint (soft nudge)
QRS div by 5 means S ends in 0 or 5; the only digit available is 5. So S = 5.
Show hint (sharpest)
PQR div by 4 means QR is divisible by 4. With S = 5 used, QR is a 2-digit from {1, 2, 3, 4}. Options: 12, 24, 32.
Show solution
  1. S = 5 (the only digit from {1,2,3,4,5} that makes QRS end in 0 or 5).
  2. QR must be a 2-digit number from {1, 2, 3, 4} divisible by 4: {12, 24, 32}.
  3. Test each. QR = 12: leftover {3, 4} for P, T. RST = 25T, digit sum 7 + T; T ∈ {3, 4} gives 10 or 11, neither div by 3.
  4. QR = 24: leftover {1, 3}. RST = 45T, digit sum 9 + T; T = 3 gives 12 (div 3) ✓. So Q = 2, R = 4, T = 3, P = 1. Number: 12435.
  5. QR = 32: leftover {1, 4}. RST = 25T sum 7 + T; T ∈ {1, 4} gives 8 or 11, neither div by 3.
A P = 1.
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Problem 25 · 2001 AMC 8 Stretch
Number Theory divisibility careful-counting

There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5, and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?

Show hint (soft nudge)
The multiple still uses the digits 2, 4, 5, 7, so it stays between 2457 and 7542.
Show hint (sharpest)
That keeps the factor tiny: 2457 × 4 is already too big, so test multiplying by 3.
Show solution
  1. Any such multiple uses the same four digits, so it lies between 2457 and 7542; since 2457 × 4 ≈ 9828 overshoots, the factor can only be 2 or 3.
  2. No doubling of a number in the set lands back in the set, but 2475 × 3 = 7425, which uses exactly 2, 4, 5, 7.
  3. So the number is 7425 (= 3 × 2475).
D 7425.
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Problem 24 · 1999 AMC 8 Stretch
Number Theory units-digit cyclicity

When 19992000 is divided by 5, the remainder is

Show hint (soft nudge)
Only the units digit of 1999 matters for the units digit of the power.
Show hint (sharpest)
Powers of 9 end in 9, 1, 9, 1, … — and an even exponent lands on 1.
Show solution
  1. Powers of a number ending in 9 cycle 9, 1, 9, 1, …; since 2000 is even, 1999²⁰⁰⁰ ends in 1.
  2. A number ending in 1 leaves remainder 1 when divided by 5.
D 1.
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Problem 24 · 1998 AJHSME Stretch
Number Theory triangular-numbers mod-arithmetic

A board of 8 columns has squares numbered left to right, top to bottom (row one is 1–8, row two is 9–16, and so on). A student shades square 1, then skips one and shades square 3, skips two and shades square 6, skips three and shades square 10, and continues this way until every column has at least one shaded square. What is the number of the shaded square that first achieves this?

Show hint (soft nudge)
The shaded squares are the triangular numbers 1, 3, 6, 10, 15, … .
Show hint (sharpest)
A square's column is set by its value mod 8 — you need all 8 remainders to appear.
Show solution
  1. The shaded squares are triangular numbers, and a square's column depends on its value mod 8.
  2. Every remainder shows up early except 0; the first triangular number divisible by 8 is 120, so the last column fills at square 120.
E 120.
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Problem 23 · 1997 AJHSME Stretch
Number Theory bounding casework

Some positive integers have both properties: (I) the sum of the squares of their digits is 50, and (II) each digit is larger than the one to its left. The product of the digits of the largest such integer is

Show hint (soft nudge)
The digits strictly increase, so they're distinct; squaring shows there can be at most four of them.
Show hint (sharpest)
To make the number large, push the leading digits up while keeping the square-sum at 50.
Show solution
  1. Five increasing digits would have squares summing to at least 1+4+9+16+25 = 55 > 50, so at most four digits.
  2. The digits 1, 2, 3, 6 give 1 + 4 + 9 + 36 = 50 and form the largest such number, 1236.
  3. Its digit product is 1 × 2 × 3 × 6 = 36.
C 36.
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Problem 25 · 1997 AJHSME Stretch
Number Theory units-digit cyclicity

All the even numbers from 2 to 98 inclusive, except those ending in 0, are multiplied together. What is the units digit of the product?

Show hint (soft nudge)
Only units digits matter; each block of ten contributes the units 2, 4, 6, 8.
Show hint (sharpest)
2 × 4 × 6 × 8 ends in 4, and there are ten blocks — find the units digit of 4¹⁰.
Show solution
  1. Each block of ten (like 2, 4, 6, 8) multiplies to a units digit of 4, and there are 10 such blocks, so we need the units digit of 4¹⁰.
  2. 4² ends in 6, and any power of 6 ends in 6, so 4¹⁰ = (4²)⁵ ends in 6.
D 6.
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Problem 23 · 1994 AJHSME Stretch
Number Theory place-value maximize
ajhsme-1994-23
Show hint (soft nudge)
The sum equals 113·X + 10·Y; keep it to three digits while making it large.
Show hint (sharpest)
Find the digits of that largest sum and match them to X, Y, Z.
Show solution
  1. Adding XXX + YX + X gives 113·X + 10·Y. To stay three digits, X ≤ 8; taking X = 8, Y = 9 gives 904 + 90 = 994.
  2. 994 reads as 9, 9, 4 = Y, Y, (a new digit Z), so the form is YYZ.
D Form YYZ.
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Problem 25 · 1994 AJHSME Stretch
Number Theory small-cases pattern
ajhsme-1994-25
Show hint (soft nudge)
Try small versions: 9 × 4, 99 × 44, 999 × 444, and find the digit-sum pattern.
Show hint (sharpest)
The digit sum turns out to be 9 times the number of nines.
Show solution
  1. 9·4 = 36 (digit sum 9), 99·44 = 4356 (18), 999·444 = 443556 (27): the digit sum is 9 × (number of nines).
  2. With 94 nines, the digit sum is 9 × 94 = 846.
A 846.
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Problem 20 · 1993 AJHSME Stretch
Number Theory borrowing-pattern digit-sum

When 1093 − 93 is expressed as a single whole number, the sum of the digits is

Show hint (soft nudge)
Try a small case like 10⁴ − 93 = 9907 to see the shape.
Show hint (sharpest)
10ⁿ − 93 is a string of nines ending in 07.
Show solution
  1. 10⁴ − 93 = 9907, 10⁵ − 93 = 99907 — so 10ⁿ − 93 is (n − 2) nines followed by 07.
  2. For n = 93 that's 91 nines and a 0 and 7: 91 × 9 + 7 = 826.
D 826.
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Problem 24 · 1993 AJHSME Stretch
Number Theory pattern rows
ajhsme-1993-24
Show hint (soft nudge)
Row k ends at the perfect square k², and holds 2k − 1 numbers.
Show hint (sharpest)
Find which row holds 142, then the number sitting one row up and aligned with it.
Show solution
  1. Rows end at 1, 4, 9, 16, …, k², so 142 is in row 12 (122–144), as its 21st of 23 entries.
  2. Row 11 (101–121) sits centered above, so directly above the 21st entry is the 20th entry of row 11: 101 + 19 = 120.
C 120.
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Problem 20 · 1991 AJHSME Stretch
Number Theory cryptarithm place-value
ajhsme-1991-20
Show hint (soft nudge)
Add the column-place values: ABC + AB + A = 111·A + 11·B + C.
Show hint (sharpest)
Set that equal to 300 and find digits A, B, C.
Show solution
  1. 111A + 11B + C = 300. Taking A = 2 gives 11B + C = 78, so B = 7 and C = 1 (271 + 27 + 2 = 300).
  2. All digits differ, so C = 1.
A 1.
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Problem 11 · 1990 AJHSME Stretch
Number Theory consecutive opposite-pairs
ajhsme-1990-11
Show hint (soft nudge)
The six faces are six consecutive numbers including the visible 11, 14, 15.
Show hint (sharpest)
The visible faces meet at a corner, so they can't be opposite each other — that pins which six numbers.
Show solution
  1. The visible 11, 14, 15 are mutually adjacent, so they aren't opposite pairs. The six consecutive numbers must be 11–16, pairing as 11+16, 12+15, 13+14 (each summing to 27).
  2. Their total is 11 + 12 + 13 + 14 + 15 + 16 = 81.
E 81.
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Problem 22 · 1990 AJHSME Stretch
Number Theory modular divisors

Several students are seated at a large circular table. They pass around a bag of 100 pieces of candy. Each person takes one piece and passes the bag to the next person. If Chris takes the first and the last piece of candy, then the number of students at the table could be

Show hint (soft nudge)
Chris takes pieces 1, n+1, 2n+1, … where n is the number of students.
Show hint (sharpest)
Taking the 100th piece too means 99 is a multiple of n.
Show solution
  1. Chris takes the 1st and every nth piece after, so for the 100th to be his, n must divide 100 − 1 = 99.
  2. Of the choices, only 11 divides 99, so there could be 11 students.
B 11.
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Problem 22 · 1989 AJHSME Stretch
Number Theory lcm-of-cycle-lengths

The letters A, J, H, S, M, E and the digits 1, 9, 8, 9 are "cycled" separately as follows and put together in a numbered list:

      AJHSME  1989
  1.  JHSMEA  9891
  2.  HSMEAJ  8919
  3.  SMEAJH  9198
      .........

What is the number of the line on which AJHSME 1989 will appear for the first time?

Show hint (soft nudge)
The letters return to AJHSME every 6 lines; the digits return to 1989 every 4 lines.
Show hint (sharpest)
The two parts line up again the first time both cycles finish together.
Show solution
  1. The 6 letters come back after 6 cycles; the 4 digits come back after 4 cycles.
  2. Both line up together at the smallest common multiple: LCM(6, 4) = 12.
C 12.
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Problem 25 · 1986 AJHSME Stretch
Number Theory average-of-arithmetic-progression

Which of the following sets of whole numbers has the largest average?

Show hint (soft nudge)
For an arithmetic progression, the average is just (first + last)⁄2.
Show hint (sharpest)
Compare each set's first and last terms.
Show solution
  1. Averages: multiples of 2 → (2 + 100)⁄2 = 51; of 3 → (3 + 99)⁄2 = 51; of 4 → (4 + 100)⁄2 = 52; of 5 → (5 + 100)⁄2 = 52.5; of 6 → (6 + 96)⁄2 = 51.
  2. Largest is 52.5 — multiples of 5.
D multiples of 5 between 1 and 101.
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