Topic

Algebra & Patterns

Formulas, made-up operations, sequences.

107 problems
Practice
Problem 4 · 2025 AMC 8 Medium
Algebra & Patterns arithmetic-sequence off-by-one

Lucius is counting backward by 7s. His first three numbers are 100, 93, and 86. What is his 10th number?

Show hint (soft nudge)
From the 1st number to the 10th, how many steps do you actually take?
Show hint (sharpest)
From the 1st number to the 10th you take 9 steps of 7. How much do you subtract in total?
Show solution
  1. Each step subtracts 7, and from the 1st to the 10th number is 9 steps.
  2. Total subtracted: 9 × 7 = 63.
  3. 10th number: 100 − 63 = 37.
B 37.
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Problem 3 · 2023 AMC 8 Medium
Algebra & Patterns evaluate-formula

Wind chill estimates how cold it feels in wind, using

(wind chill) = (air temperature) − 0.7 × (wind speed),

with temperature in °F and wind speed in mph. If the air temperature is 36°F and the wind speed is 18 mph, which is closest to the wind chill?

Show hint
Just put the numbers into the formula. Do the multiplication first, then subtract.
Show solution
  1. Multiply: 0.7 × 18 = 12.6.
  2. Subtract from the temperature: 36 − 12.6 = 23.4.
  3. The closest answer choice is 23.
B About 23.
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Problem 2 · 2022 AMC 8 Easy
Algebra & Patterns custom-operation substitution

Consider these two operations:

ab = a2b2
ab = (ab)2

What is the value of (5 ◆ 3) ★ 6?

Show hint
Work the inside (the ◆) first, then apply the ★ to that result.
Show solution
  1. Inside first: 5 ◆ 3 = 52 − 32 = 25 − 9 = 16.
  2. Now apply ★: 16 ★ 6 = (16 − 6)2 = 102.
  3. = 100.
D 100.
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Problem 4 · 2020 AMC 8 Medium
Algebra & Patterns arithmetic-sequence spiral-pattern
amc8-2020-04
Show hint (soft nudge)
Don't count the next picture — count what each new band adds to the one before it.
Show hint (sharpest)
Each new band of a hexagon has 6 more dots than the previous band: 6, 12, 18, … Add the next one to the 19 you already have.
Show solution
  1. Each new band adds 6 more dots than the last: bands are 6, 12, 18, 24, …
  2. The 3rd hexagon already has 1 + 6 + 12 = 19 dots.
  3. The next hexagon adds the 18-dot band: 19 + 18 = 37.
B 37 dots.
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Problem 7 · 2014 AMC 8 Easy
Algebra & Patterns sum-and-difference

There are four more girls than boys in Ms. Raub's class of 28 students. What is the ratio of number of girls to the number of boys in her class?

Show hint
Sum is 28 and difference is 4. The larger group is (sum + diff)/2, the smaller is (sum − diff)/2.
Show solution
  1. Girls = (28 + 4)/2 = 16, boys = (28 − 4)/2 = 12.
  2. Ratio = 16 : 12 = 4 : 3.
B 4 : 3.
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Problem 4 · 2013 AMC 8 Easy
Algebra & Patterns find-share-then-total

Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?

Show hint
The 7 friends together covered Judi's share: 7 × $2.50. That's one-eighth of the total.
Show solution
  1. Judi's share = 7 × $2.50 = $17.50.
  2. Everyone paid the same, so total = 8 × $17.50 = $140.
C $140.
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Problem 9 · 2013 AMC 8 Easy
Algebra & Patterns powers-of-two

The Incredible Hulk can double the distance it jumps with each succeeding jump. If its first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will it first be able to jump more than 1 kilometer (1,000 meters)?

Show hint
Jump n is 2n−1 meters. Find the smallest n with 2n−1 > 1000.
Show solution
  1. Jump n = 2n−1 m.
  2. 29 = 512 (10th jump), 210 = 1024 (11th jump).
  3. First > 1000 is the 11th.
C 11th jump.
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Problem 17 · 2013 AMC 8 Easy
Algebra & Patterns consecutive-integers

The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

Show hint
Sum of 6 consecutive integers starting at x is 6x + 15.
Show solution
  1. Let smallest = x. Then 6x + 15 = 2013 ⇒ 6x = 1998 ⇒ x = 333.
  2. Largest = 333 + 5 = 338.
B 338.
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Problem 9 · 2012 AMC 8 Easy
Algebra & Patterns system-of-equations head-leg-trick

The Fort Worth Zoo has a number of two-legged birds and a number of four-legged mammals. On one visit to the zoo, Margie counted 200 heads and 522 legs. How many of the animals that Margie counted were two-legged birds?

Show hint
If every animal had 4 legs, you'd see 800 legs. The shortage (800 − 522) comes from the 2-leg birds, each missing 2 legs.
Show solution
  1. If all 200 had 4 legs: 200 · 4 = 800 legs.
  2. Actual: 522 ⇒ shortage = 800 − 522 = 278 legs.
  3. Each bird is short 2 legs ⇒ birds = 278 / 2 = 139.
C 139 birds.
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Problem 19 · 2012 AMC 8 Easy
Algebra & Patterns all-but-trick

In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?

Show hint
"All but 6 are red" means green + blue = 6. Get all three two-color sums and add them up.
Show solution
  1. Green + Blue = 6, Red + Blue = 8, Red + Green = 4.
  2. Add: 2(Red + Green + Blue) = 18 ⇒ Total = 9.
C 9 marbles.
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Problem 2 · 2010 AMC 8 Easy
Algebra & Patterns operator-definition

If a@b = a × ba + b for a, b positive integers, then what is 5@10?

Show hint
Just substitute.
Show solution
  1. 5@10 = (5 · 10)/(5 + 10) = 50/15 = 10/3.
D 10/3.
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Problem 11 · 2010 AMC 8 Easy
Algebra & Patterns ratio-and-difference

The top of one tree is 16 feet higher than the top of another tree. The heights of the two trees are in the ratio 3 : 4. In feet, how tall is the taller tree?

Show hint
The ratio 3:4 means the difference (1 part) corresponds to 16 feet. So 1 part = 16 ft.
Show solution
  1. 4 − 3 = 1 part = 16 ft.
  2. Taller (4 parts) = 4 · 16 = 64 ft.
B 64 feet.
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Problem 1 · 2009 AMC 8 Easy
Algebra & Patterns work-backwards

Bridget bought a bag of apples at the grocery store. She gave half of the apples to Ann. Then she gave Cassie 3 apples, keeping 4 apples for herself. How many apples did Bridget buy?

Show hint
Work backwards from the 4 she kept.
Show solution
  1. After giving Cassie 3: she had 4 + 3 = 7.
  2. Before giving Ann half: original = 2 · 7 = 14.
E 14 apples.
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Problem 5 · 2009 AMC 8 Easy
Algebra & Patterns tribonacci recurrence

A sequence of numbers starts with 1, 2, and 3. The fourth number of the sequence is the sum of the previous three numbers in the sequence: 1 + 2 + 3 = 6. In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence?

Show hint
Just iterate the rule: each new term = sum of the previous three.
Show solution
  1. 1, 2, 3, 6, 11, 20, 37, 68.
D 68.
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Problem 19 · 2007 AMC 8 Easy
Algebra & Patterns difference-of-squares

Pick two consecutive positive integers whose sum is less than 100. Square both of those integers and then find the difference of the squares. Which of the following could be the difference?

Show hint
(x+1)2x2 = 2x + 1 = sum of the two integers. The sum < 100 and is odd.
Show solution
  1. (x+1)2x2 = 2x + 1 = (x) + (x+1).
  2. Difference equals the sum of the two integers — less than 100, and odd.
  3. Only 79 is odd and less than 100.
C 79.
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Problem 9 · 2006 AMC 8 Easy
Algebra & Patterns telescoping-product

What is the product of 32 × 43 × 54 × … × 20062005?

Show hint
Each numerator cancels with the next denominator. Only the first denominator and last numerator survive.
Show solution
  1. Product collapses to 2006 / 2 = 1003.
C 1003.
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Problem 10 · 2006 AMC 8 Easy
Algebra & Patterns inverse-proportion

Jorge's teacher asks him to plot all the ordered pairs (w, l) of positive integers for which w is the width and l is the length of a rectangle with area 12. What should his graph look like?

Show hint
wl = 12 means l = 12/w: as w increases, l decreases hyperbolically.
Show solution
  1. l = 12/w: w and l are inversely proportional — plot decreases.
  2. Only graph A shows this shape.
A Graph A.
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Problem 12 · 2005 AMC 8 Easy
Algebra & Patterns arithmetic-sequence average

Big Al the ape ate 100 delicious yellow bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many delicious bananas did Big Al eat on May 5?

Show hint
Five-term arithmetic sequence with common difference 6. Middle term = average = 100/5 = 20.
Show solution
  1. Mean = 100 / 5 = 20 = May 3.
  2. May 5 = 20 + 2 · 6 = 32.
D 32.
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Problem 4 · 2003 AMC 8 Easy
Algebra & Patterns substitution

A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted 7 children and 19 wheels. How many tricycles were there?

Show hint
Imagine all 7 children on bicycles first, then see how many extra wheels you still need.
Show solution
  1. If all 7 rode bicycles, that's 7 × 2 = 14 wheels.
  2. There are 19 − 14 = 5 extra wheels, and each tricycle adds exactly one extra wheel, so there are 5 tricycles.
C 5 tricycles.
Another way: solve the system
  1. With b bicycles and t tricycles: b + t = 7 and 2b + 3t = 19.
  2. Subtract twice the first from the second: t = 19 − 14 = 5.
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Problem 2 · 2001 AMC 8 Easy
Algebra & Patterns factor-pairs sum-constraint

I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number?

Show hint
List the factor pairs of 24 and check which pair adds to 11.
Show solution
  1. The factor pairs of 24 are 1·24, 2·12, 3·8, 4·6 — only 3 and 8 add up to 11.
  2. The larger of the two is 8.
D 8.
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Problem 1 · 1998 AJHSME Easy
Algebra & Patterns substitution

For x = 7, which of the following is the smallest?

Show hint (soft nudge)
Plug in x = 7 and compare the five values.
Show hint (sharpest)
The smallest fraction has the largest bottom and a small top.
Show solution
  1. With x = 7 the choices are 6/7, 6/8, 6/6, 7/6, 8/6.
  2. The smallest is 6/8 = 0.75, which is 6/(x+1).
B 6/(x+1).
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Problem 2 · 1998 AJHSME Easy
Algebra & Patterns custom-operation
ajhsme-1998-02
Show hint (soft nudge)
Apply the given rule: top-left × bottom-right minus top-right × bottom-left.
Show hint (sharpest)
Just plug in 3, 4, 1, 2.
Show solution
  1. The rule gives a·d − b·c = 3·2 − 4·1.
  2. That is 6 − 4 = 2.
E 2.
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Problem 2 · 1997 AJHSME Easy
Algebra & Patterns optimization

Ahn chooses a two-digit integer, subtracts it from 200, and doubles the result. What is the largest number Ahn can get?

Show hint (soft nudge)
To make the result big, subtract as little as possible from 200.
Show hint (sharpest)
The smallest two-digit number is 10.
Show solution
  1. Subtracting the smallest two-digit number, 10, leaves the most: 200 − 10 = 190.
  2. Doubling gives 2 × 190 = 380.
D 380.
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Problem 2 · 1995 AJHSME Easy
Algebra & Patterns work-through-relations

Jose is 4 years younger than Zack. Zack is 3 years older than Inez. Inez is 15 years old. How old is Jose?

Show hint (soft nudge)
Start from Inez and work outward.
Show hint (sharpest)
Zack is 3 more than Inez; Jose is 4 less than Zack.
Show solution
  1. Zack is 15 + 3 = 18.
  2. Jose is 18 − 4 = 14.
C 14.
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Problem 12 · 2024 AMC 8 Medium
Algebra & Patterns substitution sum-constraint

Rohan keeps a total of 90 guppies in 4 fish tanks.

  • There is 1 more guppy in the 2nd tank than in the 1st tank.
  • There are 2 more guppies in the 3rd tank than in the 2nd tank.
  • There are 3 more guppies in the 4th tank than in the 3rd tank.

How many guppies are in the 4th tank?

Show hint (soft nudge)
Stack the increments from tank 1: tank 2 = tank 1 + 1, tank 3 = tank 1 + 3, tank 4 = tank 1 + 6.
Show hint (sharpest)
Add all four expressions: 4·(tank 1) + 10 = 90, so tank 1 = 20.
Show solution
  1. Let tank 1 = x. Then tank 2 = x + 1, tank 3 = x + 3, tank 4 = x + 6.
  2. Sum: 4x + 10 = 90, so x = 20.
  3. Tank 4 = 20 + 6 = 26.
E 26 guppies.
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Problem 13 · 2022 AMC 8 Medium
Algebra & Patterns substitution sum-constraint

How many positive integers can fill the blank in the sentence below?

"One positive integer is ___ more than twice another, and the sum of the two numbers is 28."

Show hint (soft nudge)
Set up: smaller = a, larger = 2a + c. The blank is c.
Show hint (sharpest)
a + (2a + c) = 28 ⇒ c = 28 − 3a. For both to be positive integers, count valid a.
Show solution
  1. Let smaller = a, larger = 2a + c (with c ≥ 1 the blank). Sum: 3a + c = 28, so c = 28 − 3a.
  2. a ≥ 1 and c ≥ 1 ⇒ 28 − 3a ≥ 1 ⇒ a ≤ 9.
  3. a ∈ {1, 2, …, 9} ⇒ 9 values for the blank.
D 9 values.
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Problem 20 · 2022 AMC 8 Hard
Algebra & Patterns substitution sum-constraint
amc8-2022-20
Show hint (soft nudge)
Find the common row/column sum from the known top row, then express each missing cell in terms of x.
Show hint (sharpest)
Top row sums to 12 ⇒ magic sum is 12. The other three missing cells are 14−x, 4−x, x−1. x must beat all of them.
Show solution
  1. Top row: −2 + 9 + 5 = 12. So every row and column sums to 12.
  2. First column: −2 + (above x) + x = 12 ⇒ cell above x = 14 − x.
  3. Bottom row: x + (middle bottom) + 8 = 12 ⇒ middle bottom = 4 − x.
  4. Middle row: (14−x) + (center) + (−1) = 12 ⇒ center = x − 1.
  5. x must be the largest: x > 14−xx > 7; x > 4−xx > 2; x > x−1 always.
  6. Smallest integer satisfying all: x = 8.
D x = 8.
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Problem 16 · 2020 AMC 8 Medium
Algebra & Patterns sum-constraint substitution
amc8-2020-16
Show hint (soft nudge)
Each digit is added once per line it sits on. Which point appears on more than 2 lines?
Show hint (sharpest)
Every point is on 2 lines except B, which is on 3. So the total of all five line-sums is 2(A+B+C+D+E+F) + B = 47.
Show solution
  1. Counting incidences: each of A, C, D, E, F sits on 2 lines; B sits on 3. So the total of five line-sums is 2(A+B+C+D+E+F) + B = 47.
  2. A+B+C+D+E+F = 1+2+3+4+5+6 = 21, so 2(21) + B = 42 + B = 47.
  3. B = 5.
E B = 5.
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Problem 20 · 2020 AMC 8 Hard
Algebra & Patterns casework substitution
amc8-2020-20
Show hint (soft nudge)
Tree 2 = 11. Tree 1 and Tree 3 must each be either 22 or 5.5 (not integer) — so both are 22.
Show hint (sharpest)
Tree 4 = 22 or 44; Tree 5 = double-or-half of Tree 4. Try cases until the average ends in .2.
Show solution
  1. Tree 2 = 11. Half of 11 is 5.5 (not integer), so Tree 1 and Tree 3 must each be 22. (Tree 1 = 22, Tree 3 = 22.)
  2. Tree 4 is twice or half of Tree 3 (= 22): so Tree 4 = 44 or 11. Then Tree 5 follows from Tree 4.
  3. Cases: (T4, T5) = (44, 88): avg = 187/5 = 37.4. (44, 22): avg = 121/5 = 24.2. (11, 22): avg = 17.6. (11, 5.5) fails (not integer).
  4. Only one ends in .2: average = 24.2 meters.
B 24.2 meters.
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Problem 20 · 2019 AMC 8 Medium
Algebra & Patterns difference-of-squares factorization

How many different real numbers x satisfy the equation

(x2 − 5)2 = 16 ?
Show hint (soft nudge)
Take the square root of both sides: x2 − 5 = ±4. Two cases for x2.
Show hint (sharpest)
x2 = 9 gives two values; x2 = 1 gives two more.
Show solution
  1. (x2 − 5)2 = 16 ⇒ x2 − 5 = ±4.
  2. Case +4: x2 = 9 ⇒ x = ±3.
  3. Case −4: x2 = 1 ⇒ x = ±1.
  4. Total: 4 distinct real solutions.
D 4 real numbers.
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Problem 17 · 2017 AMC 8 Easy
Algebra & Patterns substitution sum-constraint

Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over. How many gold coins did I have?

Show hint
Set up two equations for the same gold-coin count: g = 9(n − 2) and g = 6n + 3.
Show solution
  1. Let n = chests, g = coins. Nine-per-chest leaves 2 empty: g = 9(n − 2). Six-per-chest with 3 left over: g = 6n + 3.
  2. 9(n − 2) = 6n + 3 ⇒ 3n = 21 ⇒ n = 7.
  3. g = 6(7) + 3 = 45.
C 45 coins.
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Problem 21 · 2010 AMC 8 Hard
Algebra & Patterns work-backwards

Hui is an avid reader. She bought a copy of the best seller Math is Beautiful. On the first day, Hui read 1/5 of the pages plus 12 more, and on the second day she read 1/4 of the remaining pages plus 15 pages. On the third day she read 1/3 of the remaining pages plus 18 pages. She then realized that there were only 62 pages left to read, which she read the next day. How many pages are in this book?

Show hint (soft nudge)
Work backwards from the 62 pages left. Day 3 left her with 62 pages after reading (1/3)R + 18 from R; so 2R/3 − 18 = 62.
Show hint (sharpest)
Repeat the inversion for day 2 and day 1.
Show solution
  1. After day 3 there are 62 pages left. If R3 = pages at start of day 3: (2/3)R3 − 18 = 62 ⇒ R3 = 120.
  2. Start of day 2: (3/4)R2 − 15 = 120 ⇒ R2 = 180.
  3. Start of day 1: (4/5)R1 − 12 = 180 ⇒ R1 = 240.
C 240 pages.
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Problem 24 · 2009 AMC 8 Hard
Algebra & Patterns cryptarithm

The letters A, B, C, and D represent digits. If AB + CA = DA and ABCA = A, what digit does D represent?

Show hint (soft nudge)
From the units column of the addition (B + A ends in A), get B = 0.
Show hint (sharpest)
Then AB = 10A. Plug into the subtraction equation to find A and C.
Show solution
  1. Units of addition: B + AA (mod 10) ⇒ B = 0 (no carry since A is a digit).
  2. Now AB = 10A and CA = 10C + A. Subtraction: 10A − (10C + A) = A ⇒ 9A − 10C = A ⇒ 8A = 10C ⇒ 4A = 5C.
  3. Digits with 4A = 5C: A = 5, C = 4.
  4. D · 10 + A = AB + CA = 50 + 45 = 95 ⇒ D = 9.
E 9.
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Problem 17 · 2002 AMC 8 Hard
Algebra & Patterns work-backward

In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have?

Show hint (soft nudge)
Imagine she got all 10 right first, then see what each wrong answer costs.
Show hint (sharpest)
Turning one correct answer into a wrong one drops the score by 5 + 2 = 7.
Show solution
  1. All 10 correct would score 5 × 10 = 50.
  2. Each wrong answer (instead of right) costs 5 + 2 = 7 points, and 50 − 29 = 21 = 3 × 7.
  3. So 3 were wrong and 7 were correct.
C 7.
Another way: set up an equation
  1. Let x = number correct, so 10 − x are wrong: 5x − 2(10 − x) = 29.
  2. Then 7x − 20 = 29, giving x = 7.
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Problem 17 · 2000 AMC 8 Hard
Algebra & Patterns custom-operation

The operation ★ is defined for all nonzero numbers by ab = a2b. Determine [(1 ★ 2) ★ 3] − [1 ★ (2 ★ 3)].

Show hint (soft nudge)
Compute each bracket separately, innermost operation first.
Show hint (sharpest)
The operation isn't associative, so the two brackets won't match.
Show solution
  1. 1 ★ 2 = 1²/2 = ½, then (½) ★ 3 = (½)²/3 = 1/12.
  2. 2 ★ 3 = 2²/3 = 4/3, then 1 ★ (4/3) = 1²/(4/3) = 3/4.
  3. Difference = 1/12 − 3/4 = 1/12 − 9/12 = −2/3.
A −2/3.
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Problem 19 · 1997 AJHSME Hard
Algebra & Patterns telescoping
ajhsme-1997-19
Show hint (soft nudge)
Each numerator cancels the denominator of the next fraction.
Show hint (sharpest)
After all the cancellation only a/2 survives, and b is one less than a.
Show solution
  1. Everything cancels except a/2, so a/2 = 9 gives a = 18, and b = 17.
  2. Their sum is 18 + 17 = 35.
D 35.
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Problem 16 · 1996 AJHSME Hard
Algebra & Patterns grouping telescoping

1 − 2 − 3 + 4 + 5 − 6 − 7 + 8 + 9 − 10 − 11 + … + 1992 + 1993 − 1994 − 1995 + 1996 =

Show hint (soft nudge)
The signs repeat in a 4-term pattern: +, −, −, +.
Show hint (sharpest)
Group the terms in blocks of four and see what each block adds to.
Show solution
  1. Each block of four, like 1 − 2 − 3 + 4, adds to 0.
  2. Since 1996 is a multiple of 4, the whole sum splits into 499 such blocks, giving 0.
C 0.
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Problem 20 · 1996 AJHSME Hard
Algebra & Patterns find-the-cycle

A special key on a calculator replaces the displayed number x with 1 ÷ (1 − x). (For example, from 2 it gives 1 ÷ (1 − 2) = −1.) If the calculator shows 5 and the key is pressed 100 times in a row, the calculator will display

Show hint (soft nudge)
Press the key a few times and watch for a repeating cycle.
Show hint (sharpest)
Once you know the cycle length, reduce 100 by it.
Show solution
  1. Starting at 5: 5 → −0.25 → 0.8 → 5, a cycle of length 3.
  2. Since 100 = 3·33 + 1, after 100 presses the display matches one press: −0.25.
A −0.25.
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Problem 23 · 1996 AJHSME Hard
Algebra & Patterns system-equations

The manager of a company planned to give a $50 bonus to each employee from the company fund, but the fund was $5 short of what was needed. Instead the manager gave each employee a $45 bonus and kept the remaining $95 in the fund. How much money was in the company fund before any bonuses were paid?

Show hint (soft nudge)
Let n be the number of employees and write the fund two ways.
Show hint (sharpest)
Fund = 50n − 5 (just short of $50 each) and fund = 45n + 95 (after the $45 bonuses).
Show solution
  1. The fund is 50n − 5 (5 short of $50 each) and also 45n + 95 (gave $45 each, kept $95).
  2. Setting them equal: 50n − 5 = 45n + 95 gives n = 20, so the fund was 45·20 + 95 = $995.
E 995 dollars.
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Problem 17 · 1994 AJHSME Hard
Algebra & Patterns partial-sums

Pauline can shovel snow at the rate of 20 cubic yards for the first hour, 19 cubic yards for the second, 18 for the third, and so on, always shoveling one cubic yard less per hour than the previous hour. If her driveway is 4 yards wide, 10 yards long, and covered with snow 3 yards deep, then the number of hours it will take her to shovel it clean is closest to

Show hint (soft nudge)
First find the total volume of snow: 4 × 10 × 3.
Show hint (sharpest)
Add 20 + 19 + 18 + … until you reach that volume.
Show solution
  1. The snow is 4 × 10 × 3 = 120 cubic yards.
  2. Running totals: 20, 39, 57, 74, 90, 105, 119 — after 7 hours she's at 119, just shy of 120, so the time is closest to 7 hours.
D 7.
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Problem 18 · 1994 AJHSME Hard
Algebra & Patterns interpret-graph
ajhsme-1994-18
Show hint (soft nudge)
Distance from home rises while going out, stays flat while shopping, then falls coming back.
Show hint (sharpest)
Because the speed changes (gentle in city, steep on highway), each side of the graph bends rather than staying a single straight line.
Show solution
  1. Going out, distance rises gently (city) then steeply (highway), so the climb curves and gets steeper; it stays flat for the hour at the mall; coming home it falls steeply (highway) then gently (city).
  2. Graph B shows these two different slopes on each side — the straight-sided trapezoid (A) would mean a single constant speed each way.
B Graph B.
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Problem 9 · 1993 AJHSME Hard
Algebra & Patterns read-table
ajhsme-1993-09
Show hint (soft nudge)
Read each starred product straight off the table.
Show hint (sharpest)
Do the two inner operations first, then combine.
Show solution
  1. From the table, 2 ∗ 4 = 3 and 1 ∗ 3 = 3.
  2. Then 3 ∗ 3 = 4.
D 4.
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Problem 19 · 1993 AJHSME Hard
Algebra & Patterns pair-terms

(1901 + 1902 + 1903 + … + 1993) − (101 + 102 + 103 + … + 193) =

Show hint (soft nudge)
Line up the two sums term by term.
Show hint (sharpest)
Each top term is exactly 1800 more than the matching bottom term.
Show solution
  1. Both sums have 93 terms, and each top term beats its partner by 1901 − 101 = 1800.
  2. So the difference is 93 × 1800 = 167,400.
A 167,400.
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Problem 19 · 1992 AJHSME Hard
Algebra & Patterns extremal

The distance between the 5th and 26th exits on an interstate highway is 118 miles. If any two exits are at least 5 miles apart, then what is the largest number of miles there can be between two consecutive exits that are between the 5th and 26th exits?

Show hint (soft nudge)
There are 21 gaps between the 5th and 26th exits.
Show hint (sharpest)
To stretch one gap as far as possible, make all the others as small as allowed (5 miles).
Show solution
  1. There are 26 − 5 = 21 gaps. Making 20 of them the minimum 5 miles uses 100 miles.
  2. The last gap can be 118 − 100 = 18 miles.
C 18 miles.
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Problem 19 · 1991 AJHSME Hard
Algebra & Patterns minimize-rest

The average (arithmetic mean) of 10 different positive whole numbers is 10. The largest possible value of any of these numbers is

Show hint (soft nudge)
The ten numbers add to 10 × 10 = 100.
Show hint (sharpest)
To make one as big as possible, make the other nine as small as possible (and different).
Show solution
  1. The numbers total 100. The nine smallest different positive numbers are 1, 2, …, 9, summing to 45.
  2. So the largest can be 100 − 45 = 55.
C 55.
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Problem 16 · 1990 AJHSME Hard
Algebra & Patterns pair-terms

1990 − 1980 + 1970 − 1960 + … − 20 + 10 =

Show hint (soft nudge)
Pair the terms: (1990 − 1980), (1970 − 1960), … each equals 10.
Show hint (sharpest)
Don't forget the leftover +10 at the end.
Show solution
  1. Pairs (1990 − 1980), (1970 − 1960), …, (30 − 20) each give 10, and there are 99 such pairs: 990.
  2. Adding the final +10 gives 990 + 10 = 1000.
D 1000.
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Problem 17 · 1989 AJHSME Hard
Algebra & Patterns bound-the-average

The number N is between 9 and 17. The average of 6, 10, and N could be

Show hint (soft nudge)
Plug the extreme values of N into the average formula to see what range the average can land in.
Show hint (sharpest)
Then check which choice falls inside that range.
Show solution
  1. Average = (6 + 10 + N)⁄3 = (16 + N)⁄3. With 9 < N < 17, this ranges from 25⁄3 ≈ 8.3 to 33⁄3 = 11.
  2. Of the choices, only 10 sits in that range (N = 14 gives average 10).
B 10.
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Problem 18 · 1989 AJHSME Hard
Algebra & Patterns involution
ajhsme-1989-18
Show hint (soft nudge)
Try the operation once, then again — does anything change after that?
Show hint (sharpest)
The reciprocal of the reciprocal is the original number.
Show solution
  1. Press once: 32 → 1⁄32. Press again: 1⁄32 → 32.
  2. So the display reads 32 again after exactly 2 presses.
B 2.
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Problem 19 · 1988 AJHSME Hard
Algebra & Patterns arithmetic-sequence-nth-term

What is the 100th number in the arithmetic sequence: 1, 5, 9, 13, 17, 21, 25, …?

Show hint (soft nudge)
The common difference is 4; the 100th term sits 99 steps after the first.
Show hint (sharpest)
100th term = 1 + 99 × 4.
Show solution
  1. First term 1, common difference 4. The 100th term = 1 + 99 × 4 = 1 + 396.
  2. = 397.
A 397.
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Problem 15 · 1987 AJHSME Hard
Algebra & Patterns solve-linear-equation

The sale ad read: "Buy three tires at the regular price and get the fourth tire for three dollars." Sam paid 240 dollars for a set of four tires at the sale. What was the regular price of one tire?

Show hint
Subtract the $3 promo tire first, then split what's left across three tires.
Show solution
  1. Three full-price tires cost 240 − 3 = 237 dollars.
  2. Regular price = 237 ⁄ 3 = 79 dollars.
D 79 dollars.
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Problem 19 · 1987 AJHSME Hard
Algebra & Patterns repeated-squaring growth
ajhsme-1987-19
Show hint
Repeated squaring goes 2 → 4 → 16 → 256 → 65536. Watch which crosses 500.
Show solution
  1. 2 → 4 → 16 → 256 → 65536. 256 isn't yet greater than 500, but 65536 is.
  2. So 4 presses are needed.
A 4.
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Problem 11 · 1986 AJHSME Hard
Algebra & Patterns custom-operation

If A ✶ B means (A + B) ⁄ 2, then (3 ✶ 5) ✶ 8 is

Show hint (soft nudge)
Apply the inner operation first, then the outer.
Show hint (sharpest)
3 ✶ 5 is the average of 3 and 5.
Show solution
  1. 3 ✶ 5 = (3 + 5)⁄2 = 4. Then 4 ✶ 8 = (4 + 8)⁄2 = 6.
  2. = 6.
A 6.
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Problem 18 · 1985 AJHSME Hard
Algebra & Patterns interval-from-inequalities

Nine copies of a certain pamphlet cost less than $10.00 while ten copies of the same pamphlet (at the same price) cost more than $11.00. How much does one copy of this pamphlet cost?

Show hint (soft nudge)
9p < 10 and 10p > 11 give a narrow interval for p.
Show hint (sharpest)
10⁄9 ≈ 1.111 and 11⁄10 = 1.10.
Show solution
  1. 9p < 10 → p < 10⁄9 ≈ 1.111. 10p > 11 → p > 1.10. So 1.10 < p < 1.111.
  2. Only choice in range: $1.11.
E $1.11.
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Problem 14 · 2026 AMC 8 Medium
Algebra & Patterns arithmetic-sequence

Jami picked three equally spaced integers on the number line. The sum of the first and second is 40, and the sum of the second and third is 60. What is the sum of all three numbers?

Show hint (soft nudge)
For equally spaced numbers, the middle one is the average of the other two.
Show hint (sharpest)
Add the two given sums and see how many times the middle number appears.
Show solution
  1. Since the numbers are equally spaced, first + third = 2 × second. Adding the two given sums: 40 + 60 = first + 2·second + third = 4·second, so the middle is 25.
  2. The total is 3 × 25 = 75.
B 75.
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Problem 9 · 2023 AMC 8 Medium
Algebra & Patterns evaluate-formula
amc8-2023-09
Show hint (soft nudge)
Draw two horizontal lines on the graph at elevations 4 and 7. Read off the time intervals where the curve sits between them.
Show hint (sharpest)
There are three such time intervals; add their durations.
Show solution
  1. The curve sits between elevations 4 and 7 over three intervals: roughly t = 2 to 4 (2 sec), t = 6 to 10 (4 sec), and t = 12 to 14 (2 sec).
  2. Total: 2 + 4 + 2 = 8 seconds.
B 8 seconds.
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Problem 6 · 2022 AMC 8 Easy
Algebra & Patterns substitution arithmetic-sequence

Three positive integers are equally spaced on a number line. The middle number is 15 and the largest number is 4 times the smallest number. What is the smallest of these three numbers?

Show hint (soft nudge)
Equally spaced means the middle equals the average of the outer two.
Show hint (sharpest)
Let smallest = x, largest = 4x. Their average = 5x/2 = 15.
Show solution
  1. Let smallest = x, so largest = 4x. Equally spaced ⇒ middle = (x + 4x)/2 = 5x/2.
  2. Set 5x/2 = 15 ⇒ x = 6.
C 6.
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Problem 8 · 2020 AMC 8 Medium
Algebra & Patterns substitution sum-constraint

Ricardo has 2020 coins, some of which are pennies (1-cent coins) and the rest of which are nickels (5-cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least possible amounts of money that Ricardo can have?

Show hint (soft nudge)
Trade pennies for nickels: each swap adds 4 cents. So the total is a linear function of the nickel count n.
Show hint (sharpest)
Total = p + 5n = (p+n) + 4n = 2020 + 4n. With 1 ≤ n ≤ 2019, difference is 4 × (2019 − 1).
Show solution
  1. Total cents = p + 5n = (p + n) + 4n = 2020 + 4n.
  2. Constraints: p ≥ 1 and n ≥ 1 with p + n = 2020 ⇒ n ∈ [1, 2019].
  3. Maximum vs minimum total: 4(2019) − 4(1) = 4 × 2018 = 8072.
C 8072 cents.
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Problem 10 · 2016 AMC 8 Easy
Algebra & Patterns custom-operation substitution

Suppose that ab means 3ab. What is the value of x if

2 ∗ (5 ∗ x) = 1 ?
Show hint
Inner first: 5 ∗ x = 15 − x. Then plug into the outer.
Show solution
  1. 5 ∗ x = 15 − x.
  2. 2 ∗ (15 − x) = 6 − (15 − x) = x − 9.
  3. Set = 1: x = 10.
D x = 10.
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Problem 9 · 2015 AMC 8 Easy
Algebra & Patterns arithmetic-sequence arithmetic-series

On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working 20 days?

Show hint
The first 20 positive odd numbers: 1, 3, 5, …, 39. The sum of the first n odd numbers is n2.
Show solution
  1. Day k sales: 2k − 1. Days 1 to 20 sum: 1 + 3 + 5 + … + 39.
  2. Sum of the first n odd numbers is n2.
  3. Total: 202 = 400.
D 400 widgets.
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Problem 16 · 2015 AMC 8 Medium
Algebra & Patterns fraction-word-problem substitution

In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If 13 of all the ninth graders are paired with 25 of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?

Show hint (soft nudge)
Pairs are one-to-one, so the number of paired ninth-graders equals the number of paired sixth-graders. Set that up as one equation and pick a convenient size.
Show hint (sharpest)
Let there be 15 ninth-graders. Then 5 of them are paired (one-third). So 5 sixth-graders are paired too, which is two-fifths of all sixth-graders ⇒ 12 sixth-graders total.
Show solution
  1. Pairs are one-to-one, so paired ninth-graders = paired sixth-graders. Pick small numbers that make both fractions whole: ninth = 6, sixth = 5.
  2. Then paired ninth = (1/3)(6) = 2 and paired sixth = (2/5)(5) = 2. ✓
  3. Total students = 6 + 5 = 11. Buddied students = 2 + 2 = 4. Fraction = 4/11.
B 4/11.
Another way: algebra
  1. Let n = ninth-graders, s = sixth-graders. (1/3)n = (2/5)s ⇒ 5n = 6s.
  2. Buddied / total = ((1/3)n + (2/5)s) / (n + s) = (2 · (1/3)n) / (n + s) (since the two numerators are equal).
  3. Using 5n = 6s, ratio simplifies to 4/11.
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Problem 17 · 2015 AMC 8 Medium
Algebra & Patterns distance-rate-time

Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?

Show hint (soft nudge)
Set the two distance expressions equal: rush-hour distance = no-traffic distance. Convert 20 min and 12 min to hours (1/3 and 1/5).
Show hint (sharpest)
Solve s · (1/3) = (s + 18) · (1/5) for the rush-hour speed s, then plug back.
Show solution
  1. Let rush-hour speed be s mph. Distance: s · (1/3) = (s + 18) · (1/5).
  2. Multiply by 15: 5s = 3(s + 18) ⇒ 2s = 54 ⇒ s = 27.
  3. Distance = 27 · (1/3) = 9 miles.
D 9 miles.
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Problem 18 · 2015 AMC 8 Medium
Algebra & Patterns arithmetic-sequence middle-is-average

An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, 2, 5, 8, 11, 14 is an arithmetic sequence with five terms, in which the first term is 2 and the constant added is 3. Each row and each column in this 5 × 5 array is an arithmetic sequence with five terms. The square in the center is labelled X. What is the value of X?

Show hint (soft nudge)
In a 5-term arithmetic sequence the middle (3rd) term equals the average of the 1st and 5th terms.
Show hint (sharpest)
Find the middle of the top row (avg of 1 and 25), middle of the bottom row (avg of 17 and 81), then average those two.
Show solution
  1. Middle of top row = (1 + 25)/2 = 13.
  2. Middle of bottom row = (17 + 81)/2 = 49.
  3. X is the middle of the middle column, which is the average of those two: (13 + 49)/2 = 31.
B X = 31.
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Problem 20 · 2015 AMC 8 Medium
Algebra & Patterns system-of-equations parity-mod

Ralph went to the store and bought 12 pairs of socks for a total of $24. Some of the socks he bought cost $1 a pair, some of the socks he bought cost $3 a pair, and some of the socks he bought cost $4 a pair. If he bought at least one pair of each type, how many pairs of $1 socks did Ralph buy?

Show hint (soft nudge)
Let a, b, c be the counts of $1, $3, $4 pairs. Write the two equations and subtract.
Show hint (sharpest)
After subtracting, you get 2b + 3c = 12 with b, c ≥ 1. Use parity (2b is even) to pin down c.
Show solution
  1. a + b + c = 12 and a + 3b + 4c = 24.
  2. Subtract: 2b + 3c = 12. So 3c is even, meaning c is even; and 0 < c < 4 ⇒ c = 2.
  3. Then 2b = 6 ⇒ b = 3, and a = 12 − 3 − 2 = 7.
D 7 pairs.
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Problem 22 · 2014 AMC 8 Medium
Algebra & Patterns place-value-algebra

A 2-digit number is such that the product of the digits plus the sum of the digits is equal to the number. What is the units digit of the number?

Show hint
Write the number as 10a + b and translate the condition. Cancel everything you can.
Show solution
  1. Let the number be 10a + b. Condition: ab + a + b = 10a + b.
  2. Simplify: ab = 9a. Since a ≠ 0 (it's a two-digit number), divide by a: b = 9.
  3. (Any 2-digit number with units digit 9 works: 19, 29, 39, … check: 1·9 + 1 + 9 = 19. ✓)
E Units digit 9.
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Problem 15 · 2013 AMC 8 Medium
Algebra & Patterns powers-by-inspection

If 3p + 34 = 90, 2r + 44 = 76, and 53 + 6s = 1421, what is the product of p, r, and s?

Show hint (soft nudge)
Subtract the known term from each equation, then identify the power.
Show hint (sharpest)
9 = 32, 32 = 25, 1296 = 64.
Show solution
  1. 3p = 90 − 81 = 9 = 32p = 2.
  2. 2r = 76 − 44 = 32 = 25r = 5.
  3. 6s = 1421 − 125 = 1296 = 64s = 4.
  4. Product: 2 · 5 · 4 = 40.
B 40.
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Problem 16 · 2012 AMC 8 Medium
Algebra & Patterns place-value-greedy

Each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is used only once to make two five-digit numbers so that they have the largest possible sum. Which of the following could be one of the numbers?

Show hint
Higher place values dominate. To maximize the sum, put the two biggest digits in the ten-thousands place (one each), the next two in the thousands place, etc.
Show solution
  1. Pair digits by descending size for each place: {9, 8}, {7, 6}, {5, 4}, {3, 2}, {1, 0}.
  2. Each number gets one from each pair. So the digits of each number (left to right) come from {9, 8}, {7, 6}, {5, 4}, {3, 2}, {1, 0}.
  3. Only 87431 matches: 8∈{9,8}, 7∈{7,6}, 4∈{5,4}, 3∈{3,2}, 1∈{1,0}. ✓
C 87431.
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Problem 13 · 2010 AMC 8 Medium
Algebra & Patterns consecutive-integers percent-equation

The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is 30% of the perimeter. What is the length of the longest side?

Show hint
Let the smallest side be s. Then perimeter = 3s + 3. Set s = 0.3 · (3s + 3).
Show solution
  1. Sides: s, s+1, s+2. Perimeter: 3s + 3.
  2. s = 0.3(3s + 3) ⇒ s = 0.9s + 0.9 ⇒ 0.1s = 0.9 ⇒ s = 9.
  3. Longest = s + 2 = 11.
E 11 inches.
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Problem 22 · 2010 AMC 8 Medium
Algebra & Patterns place-value-difference

The hundreds digit of a three-digit number is 2 more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?

Show hint (soft nudge)
Let units = u, tens = t, hundreds = u + 2. Original − reversed simplifies to a constant.
Show hint (sharpest)
Reversing flips the hundreds and units digits. Difference = 99 · (hundreds − units).
Show solution
  1. Original − reversed = 100(u+2) + 10t + u − (100u + 10t + u+2) = 99(u+2) − 99u = 198.
  2. Units digit of 198 = 8.
E 8.
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Problem 24 · 2010 AMC 8 Medium
Algebra & Patterns match-bases-or-exponents

What is the correct ordering of the three numbers, 108, 512, and 224?

Show hint (soft nudge)
Rewrite each as a product of equal-exponent powers. E.g., 108 = 28 · 58, 224 = 28 · 48, 512 = 58 · 54.
Show hint (sharpest)
Compare pairwise using shared factors.
Show solution
  1. Compare 224 vs 108: 224 = 28 · 48, 108 = 28 · 58. Since 4 < 5: 224 < 108.
  2. Compare 108 vs 512: 108 = 44 · 58, 512 = 54 · 58. Since 44 = 256 < 625 = 54: 108 < 512.
  3. Therefore 224 < 108 < 512.
A 2^24 &lt; 10^8 &lt; 5^12.
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Problem 21 · 2009 AMC 8 Medium
Algebra & Patterns invariant-total

Andy and Bethany have a rectangular array of numbers with 40 rows and 75 columns. Andy adds the numbers in each row. The average of his 40 sums is A. Bethany adds the numbers in each column. The average of her 75 sums is B. What is the value of AB?

Show hint
Both Andy and Bethany sum the same array total. So 40A = sum of array = 75B.
Show solution
  1. Sum of all entries = 40 · A = 75 · B.
  2. A/B = 75/40 = 15/8.
D 15/8.
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Problem 23 · 2009 AMC 8 Medium
Algebra & Patterns quadratic

On the last day of school, Mrs. Awesome gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought 400 jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class?

Show hint
Each boy gets b beans, so all boys get b2. Similarly girls get g2. b2 + g2 = 394.
Show solution
  1. b = g + 2 and b2 + g2 = 400 − 6 = 394.
  2. (g + 2)2 + g2 = 394 ⇒ 2g2 + 4g + 4 = 394 ⇒ g2 + 2g − 195 = 0.
  3. (g + 15)(g − 13) = 0 ⇒ g = 13, b = 15.
  4. Total: 13 + 15 = 28.
B 28 students.
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Problem 12 · 2008 AMC 8 Medium
Algebra & Patterns geometric-sequence

A ball is dropped from a height of 3 meters. On its first bounce it rises to a height of 2 meters. It keeps falling and bouncing to 23 of the height it reached in the previous bounce. On which bounce will it rise to a height less than 0.5 meters?

Show hint
After the nth bounce, height = 3 · (2/3)n. Test small n until it drops below 1/2.
Show solution
  1. After bounce 4: 3 · (2/3)4 = 16/27 ≈ 0.593 (above 0.5).
  2. After bounce 5: 3 · (2/3)5 = 32/81 ≈ 0.395 (below 0.5).
  3. Answer: 5.
C 5th bounce.
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Problem 13 · 2008 AMC 8 Medium
Algebra & Patterns system-of-equations

Mr. Harman needs to know the combined weight in pounds of three boxes he wants to mail. However, the only available scale is not accurate for weights less than 100 pounds or more than 150 pounds. So the boxes are weighed in pairs in every possible way. The results are 122, 125 and 127 pounds. What is the combined weight in pounds of the three boxes?

Show hint
Each box is in 2 of the 3 pair-sums. Adding all three pair-sums double-counts each weight.
Show solution
  1. Sum of pair-sums: 122 + 125 + 127 = 374 = 2(a + b + c).
  2. Total: 374 / 2 = 187.
C 187 pounds.
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Problem 17 · 2008 AMC 8 Medium
Algebra & Patterns fixed-perimeter-area max-min

Ms. Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of 50 units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?

Show hint
l + w = 25, both positive integers. Area = l(25 − l); max near l = 12 or 13, min at l = 1.
Show solution
  1. Largest: l = 12, w = 13 ⇒ area 156.
  2. Smallest: l = 1, w = 24 ⇒ area 24.
  3. Difference: 156 − 24 = 132.
D 132.
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Problem 20 · 2008 AMC 8 Medium
Algebra & Patterns common-numerator smallest-integer-total

The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and 3/4 of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?

Show hint (soft nudge)
Let p be the common count passing. Boys = (3/2)p, girls = (4/3)p; total = (17/6)p.
Show hint (sharpest)
Total must be a positive integer; smallest p making it integer is p = 6.
Show solution
  1. Boys = (3/2)p, girls = (4/3)p. Total = (3/2 + 4/3)p = (17/6)p.
  2. Smallest positive integer total requires p = 6 ⇒ total = 17.
B 17.
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Problem 15 · 2007 AMC 8 Medium
Algebra & Patterns inequality-reasoning

Let a, b and c be numbers with 0 < a < b < c. Which of the following is impossible?

Show hint
a > 0 and c > b imply a + c > b.
Show solution
  1. Since c > b and a > 0, a + c > b — never less.
  2. All other choices have explicit examples (e.g., a = 1/3, b = 1/2, c = 1 gives ac = 1/3 < 1/2 = b).
  3. Impossible: (A).
A a + c &lt; b is impossible.
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Problem 16 · 2007 AMC 8 Medium
Algebra & Patterns quadratic-vs-linear

Amanda draws five circles with radii 1, 2, 3, 4 and 5. Then for each circle she plots the point (C, A), where C is its circumference and A is its area. Which of the following could be her graph?

Show hint
C = 2πr, A = πr2. Eliminating r: A = C2/(4π). Look for a graph that's an increasing concave-up curve.
Show solution
  1. A grows like C2: both increasing, concave-up.
  2. Only graph A shows that pattern.
A Graph A.
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Problem 20 · 2007 AMC 8 Medium
Algebra & Patterns percent-equation

Before district play, the Unicorns had won 45% of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?

Show hint
Let pre-district games be x. Pre-district wins: 0.45x. Final wins: 0.45x + 6 = (x + 8)/2.
Show solution
  1. 0.45x + 6 = (x + 8)/2.
  2. Multiply by 10: 4.5x + 60 = 5x + 40 ⇒ 0.5x = 20 ⇒ x = 40.
  3. Total games: 40 + 8 = 48.
A 48 games.
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Problem 22 · 2006 AMC 8 Medium
Algebra & Patterns pyramid-formula

Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell?

Show hint (soft nudge)
Top = a + 2b + c where a, b, c are bottom-left, middle, bottom-right.
Show hint (sharpest)
Use 1, 2, 3 (min) and 7, 8, 9 (max), with the largest digit in the middle.
Show solution
  1. Top = a + 2b + c. Middle has double weight.
  2. Min: a=2, b=1, c=3 ⇒ 2 + 2 + 3 = 7.
  3. Max: a=8, b=9, c=7 ⇒ 8 + 18 + 7 = 33.
  4. Diff: 33 − 7 = 26.
D 26.
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Problem 11 · 2002 AMC 8 Medium
Algebra & Patterns perfect-square difference-of-squares

A sequence of squares is made of identical square tiles. Each square's edge is one tile longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?

Show hint (soft nudge)
A square that is n tiles on a side uses n × n tiles.
Show hint (sharpest)
Compare 7 × 7 with 6 × 6.
Show solution
  1. A square that is n tiles on a side uses n² tiles.
  2. So the seventh needs 7² = 49 and the sixth needs 6² = 36: the difference is 49 − 36 = 13.
C 13.
Another way: difference of squares (no squaring needed)
  1. 7² − 6² = (7 + 6)(7 − 6) = 13 × 1 = 13.
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Problem 12 · 2001 AMC 8 Medium
Algebra & Patterns custom-operation

If ab = a + bab, then (6 ♦ 4) ♦ 3 =

Show hint (soft nudge)
Work inside the parentheses first, just like any operation.
Show hint (sharpest)
Order matters: the bottom is a − b, not b − a.
Show solution
  1. First 6 ♦ 4 = (6 + 4)/(6 − 4) = 10/2 = 5.
  2. Then 5 ♦ 3 = (5 + 3)/(5 − 3) = 8/2 = 4.
A 4.
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Problem 12 · 1998 AJHSME Medium
Algebra & Patterns simplify-term

What is the value of 2(1 − 12) + 3(1 − 13) + 4(1 − 14) + … + 10(1 − 110)?

Show hint (soft nudge)
Simplify a single term: k(1 − 1/k) is just k − 1.
Show hint (sharpest)
Then you're adding 1 + 2 + … + 9.
Show solution
  1. Each term k(1 − 1/k) equals k − 1, so the sum is 1 + 2 + 3 + … + 9.
  2. That total is 45.
A 45.
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Problem 6 · 1996 AJHSME Medium
Algebra & Patterns minimize

What is the smallest result that can be obtained from the following process? Choose three different numbers from the set {3, 5, 7, 11, 13, 17}, add two of them, then multiply their sum by the third number.

Show hint (soft nudge)
To make a product small, multiply by the smallest number.
Show hint (sharpest)
Then add the two next-smallest numbers.
Show solution
  1. Use 3 as the multiplier and add the next two smallest, 5 and 7: (5 + 7) × 3 = 36.
  2. Any other choice (such as (3 + 7) × 5 = 50) is larger, so the smallest is 36.
C 36.
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Problem 9 · 1996 AJHSME Medium
Algebra & Patterns reciprocal

If 5 times a number is 2, then 100 times the reciprocal of the number is

Show hint (soft nudge)
First find the number, then flip it.
Show hint (sharpest)
5 × number = 2 means the number is 2/5.
Show solution
  1. The number is 2/5, so its reciprocal is 5/2.
  2. 100 × (5/2) = 250.
D 250.
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Problem 11 · 1996 AJHSME Medium
Algebra & Patterns compare-magnitudes

Let x be the number 0.00…01, where there are 1996 zeros after the decimal point before the 1. Which of the following expressions represents the largest number?

Show hint (soft nudge)
x is an extremely tiny positive number.
Show hint (sharpest)
Dividing by something tiny makes the result enormous.
Show solution
  1. Since x is a tiny positive number, 3 + x, 3 − x, and 3·x are all near 3 (or near 0), and x/3 is tiny.
  2. But 3/x divides by something tiny, giving a gigantic number, so 3/x is largest.
D 3/x.
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Problem 6 · 1992 AJHSME Medium
Algebra & Patterns custom-operation
ajhsme-1992-06
Show hint (soft nudge)
Each triangle means top + bottom-left − bottom-right.
Show hint (sharpest)
Evaluate both triangles, then add.
Show solution
  1. First triangle: 1 + 3 − 4 = 0. Second triangle: 2 + 5 − 6 = 1.
  2. Their sum is 0 + 1 = 1.
D 1.
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Problem 8 · 1992 AJHSME Medium
Algebra & Patterns cost-revenue-profit

A store owner bought 1500 pencils at $0.10 each. If he sells them for $0.25 each, how many of them must he sell to make a profit of exactly $100.00?

Show hint (soft nudge)
First find his total cost, then the revenue needed for a $100 profit.
Show hint (sharpest)
Divide that revenue by the $0.25 selling price.
Show solution
  1. Cost is 1500 × $0.10 = $150, so he needs $150 + $100 = $250 in sales.
  2. At $0.25 each, that's $250 ÷ $0.25 = 1000 pencils.
C 1000.
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Problem 4 · 1991 AJHSME Medium
Algebra & Patterns sum-near-round-number

If 991 + 993 + 995 + 997 + 999 = 5000 − N, then N =

Show hint (soft nudge)
Each term is a little under 1000; how much under?
Show hint (sharpest)
The five shortfalls add to N.
Show solution
  1. The five numbers fall short of 1000 by 9, 7, 5, 3, 1, totaling 25, so the sum is 5000 − 25.
  2. Thus N = 25.
E 25.
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Problem 21 · 1991 AJHSME Medium
Algebra & Patterns linear-rate

For every 3° rise in temperature, the volume of a certain gas expands by 4 cubic centimeters. If the volume of the gas is 24 cubic centimeters when the temperature is 32°, what was the volume in cubic centimeters when the temperature was 20°?

Show hint (soft nudge)
From 32° down to 20° is a 12° drop — how many 3° steps is that?
Show hint (sharpest)
Each step removes 4 cm³.
Show solution
  1. A 12° drop is four 3° steps, each shrinking the gas by 4 cm³, for 16 cm³ less.
  2. So the volume was 24 − 16 = 8 cm³.
A 8.
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Problem 7 · 1989 AJHSME Medium
Algebra & Patterns coin-value

If the value of 20 quarters and 10 dimes equals the value of 10 quarters and n dimes, then n =

Show hint (soft nudge)
Trading away 10 quarters loses 250 cents — make it up in dimes.
Show hint (sharpest)
Each dime is worth 10 cents.
Show solution
  1. 20 quarters + 10 dimes = 600¢. The other side is 250¢ from 10 quarters, so the dimes must supply 350¢.
  2. 350 ÷ 10 = 35 dimes.
D 35.
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Problem 7 · 1985 AJHSME Medium
Algebra & Patterns pattern-by-row
ajhsme-1985-07
Show hint (soft nudge)
In row n the row has 2n − 1 squares; counting whites first, blacks are one fewer.
Show hint (sharpest)
Row n: n whites and (n − 1) blacks.
Show solution
  1. Row 1: 1 white, 0 black. Row 2: W B W (2 white, 1 black). Row n: n whites and (n − 1) blacks.
  2. Row 37: 37 − 1 = 36 black squares.
C 36.
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Problem 20 · 2025 AMC 8 Hard
Algebra & Patterns arithmetic-series fraction-to-decimal

Sarika, Dev, and Rajiv are sharing a large block of cheese. They take turns cutting off half of what remains and eating it: first Sarika eats half of the cheese, then Dev eats half of the remaining half, then Rajiv eats half of what remains, then back to Sarika, and so on. They stop when the cheese is too small to see. About what fraction of the original block of cheese does Sarika eat in total?

Show hint (soft nudge)
Sarika eats on turns 1, 4, 7, … — every third turn. What fraction does each Sarika-bite eat?
Show hint (sharpest)
Sarika's bites are 1/2, then 1/16, then 1/128, … — a geometric series with ratio 1/8.
Show solution
  1. Each turn eats half of what's left, so the cheese remaining after turn n is 1/2n. Sarika eats at turns 1, 4, 7, …, taking 12, 116, 1128, … of the original block.
  2. Geometric series with first term a = 12 and ratio r = 18.
  3. Sum = a1 − r = 1/27/8 = 4/7.
A 4/7.
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Problem 21 · 2024 AMC 8 Hard
Algebra & Patterns ratio proportion substitution

A group of frogs (called an army) is living in a tree. A frog turns green when in the shade and turns yellow when in the sun. Initially, the ratio of green to yellow frogs was 3 : 1. Then 3 green frogs moved to the sunny side and 5 yellow frogs moved to the shady side. Now the ratio is 4 : 1. What is the difference between the number of green frogs and yellow frogs now?

Show hint (soft nudge)
Express both populations in one variable (yellow), then write the new ratio as an equation.
Show hint (sharpest)
After moves: green = 3y + 2, yellow = y − 2. Set (3y+2)/(y−2) = 4.
Show solution
  1. Let y = initial yellow count, so initial green = 3y.
  2. After the moves: green = 3y + 5 − 3 = 3y + 2 (5 in, 3 out). Yellow = y + 3 − 5 = y − 2.
  3. New ratio: 3y + 2y − 2 = 4 ⇒ 3y + 2 = 4y − 8 ⇒ y = 10. So initial green = 30, yellow = 10.
  4. After: green = 32, yellow = 8. Difference = 32 − 8 = 24.
E 24 frogs.
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Problem 25 · 2023 AMC 8 Hard
Algebra & Patterns arithmetic-sequence divisibility casework

Fifteen integers a1, a2, a3, …, a15 are arranged in order on a number line. The integers are equally spaced and have the property that

1 ≤ a1 ≤ 10,   13 ≤ a2 ≤ 20,   and   241 ≤ a15 ≤ 250.

What is the sum of the digits of a14?

Show hint (soft nudge)
Equally spaced → arithmetic sequence with common difference d. From the bounds, get a tight range for 14d.
Show hint (sharpest)
a15a1 = 14d. With 241−10 ≤ 14d ≤ 250−1, the only multiple of 14 in [231, 249] is 238 = 14×17. So d = 17.
Show solution
  1. Equal spacing: a15a1 = 14d. Bounds give 231 ≤ 14d ≤ 249, and the only multiple of 14 in there is 238 → d = 17.
  2. Then a2 = a1 + 17 forces a1 ≤ 3, and a15 = a1 + 238 forces a1 ≥ 3. So a1 = 3.
  3. a14 = a15d = (3 + 238) − 17 = 224. Digit sum: 2 + 2 + 4 = 8.
A 8.
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Problem 25 · 2020 AMC 8 Hard
Algebra & Patterns substitution sum-constraint
amc8-2020-25
Show hint (soft nudge)
Set up two equations: width = s1 + s2 + s3 and height has R2's height + s3.
Show hint (sharpest)
R2's height = s1s2, so height = s1s2 + s3. Subtract the two equations to isolate s2.
Show solution
  1. Across the top: width = s1 + s2 + s3 = 3322.
  2. Down the side: height = (height of R2) + s3. R2's height = s1s2, so height = s1s2 + s3 = 2020.
  3. Subtract: 2s2 = 3322 − 2020 = 1302 ⇒ s2 = 651.
A S2 has side 651.
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Problem 23 · 2019 AMC 8 Hard
Algebra & Patterns divisibility casework

After Euclid High School's last basketball game, it was determined that 14 of the team's points were scored by Alexa and 27 were scored by Brittany. Chelsea scored 15 points. None of the other 7 team members scored more than 2 points. What was the total number of points scored by the other 7 team members?

Show hint (soft nudge)
Let T = total. Then 1/4 and 2/7 of T are integers ⇒ T is a multiple of lcm(4,7) = 28.
Show hint (sharpest)
Others = TT/4 − 2T/7 − 15 = 13T/28 − 15. Constrained by 0 ≤ Others ≤ 7×2 = 14.
Show solution
  1. Let T be the total. Alexa has T/4, Brittany 2T/7 — both integers ⇒ T divisible by 28.
  2. Others' total: TT/4 − 2T/7 − 15 = 13T/28 − 15. Must be in [0, 14] (since 7 players, ≤ 2 each).
  3. T = 28: gives −2 (invalid). T = 56: gives 13·56/28 − 15 = 26 − 15 = 11 (valid).
  4. T = 84 would give 39 − 15 = 24, exceeding 14.
B 11 points.
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Problem 21 · 2017 AMC 8 Hard
Algebra & Patterns casework substitution

Suppose a, b, and c are nonzero real numbers, and a + b + c = 0. What are the possible value(s) for

a|a| + b|b| + c|c| + abc|abc| ?
Show hint (soft nudge)
x/|x| = +1 or −1 depending on sign. Since a + b + c = 0, exactly one or two of them are negative.
Show hint (sharpest)
Case 2 positive + 1 negative vs 1 positive + 2 negative — check the four signs sum to 0 in both cases.
Show solution
  1. Since a + b + c = 0 and none are zero, either (i) two are positive and one is negative, or (ii) two are negative and one is positive.
  2. Case (i): signs of a, b, c are +, +, −; product abc is negative. Sum: (+1) + (+1) + (−1) + (−1) = 0.
  3. Case (ii): signs are −, −, +; product is positive. Sum: (−1) + (−1) + (+1) + (+1) = 0.
  4. Either way, the value is 0.
A 0.
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Problem 23 · 2003 AMC 8 Stretch
Algebra & Patterns
amc8-2003-23
Show hint (soft nudge)
The cat and mouse move independently — handle each one's cycle on its own.
Show hint (sharpest)
The cat repeats every 4 moves and the mouse every 8; use the remainders of 247.
Show solution
  1. The cat's position repeats every 4 moves and the mouse's every 8, so only the remainder of 247 matters for each.
  2. 247 = 4·61 + 3, so the cat is where it is after 3 moves: the bottom-right square.
  3. 247 = 8·30 + 7, so the mouse is where it is after 7 moves: the bottom-left segment.
  4. The picture matching both is A.
A the cat in the bottom-right square, the mouse on the bottom-left segment.
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Problem 25 · 2002 AMC 8 Stretch
Algebra & Patterns substitution

Loki, Moe, Nick, and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money, and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?

Show hint (soft nudge)
The three gifts are equal — pick a convenient size for that common gift, like $1.
Show hint (sharpest)
Then Moe started with $5, Loki $4, Nick $3; passing money around never changes the group total.
Show solution
  1. Let each gift be $1. Since $1 is Moe's fifth, Loki's fourth, and Nick's third, they began with $5, $4, and $3.
  2. Handing money over doesn't change the total: $5 + $4 + $3 = $12, and Ott now holds the three gifts, $3.
  3. Ott's share is 3/12 = 1/4.
B 1/4.
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Problem 22 · 1998 AJHSME Stretch
Algebra & Patterns find-the-cycle

Terri builds a sequence of positive integers by these rules: if the integer is less than 10, multiply it by 9; if it is even and greater than 9, divide it by 2; if it is odd and greater than 9, subtract 5. Find the 98th term of the sequence that begins 98, 49, … .

Show hint (soft nudge)
Generate terms until they start repeating.
Show hint (sharpest)
Once you spot the repeating block, use its length to jump ahead to the 98th term.
Show solution
  1. The sequence runs 98, 49, 44, 22, 11, 6, 54, 27, 22, 11, … — from the 4th term on it cycles 22, 11, 6, 54, 27 with length 5.
  2. From the 4th term, (98 − 4) = 94 steps and 94 ÷ 5 leaves remainder 4, landing on the 5th entry of the cycle: 27.
D 27.
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Problem 23 · 1998 AJHSME Stretch
Algebra & Patterns find-the-pattern
ajhsme-1998-23
Show hint (soft nudge)
Count shaded vs. total small triangles in the first few figures, then find the rule.
Show hint (sharpest)
The ratio simplifies to (n − 1)/(2n) for the nth triangle.
Show solution
  1. The nth triangle splits into n² small triangles, of which (n² − n)/2 are shaded — a ratio of (n − 1)/(2n).
  2. For the 8th triangle that is 7/16.
C 7/16.
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Problem 25 · 1998 AJHSME Stretch
Algebra & Patterns work-backward invariant

Three generous friends redistribute their money as follows: Amy gives Jan and Toy enough to double each of their amounts; then Jan gives Amy and Toy enough to double theirs; finally Toy gives Amy and Jan enough to double theirs. Toy had $36 at the beginning and $36 at the end. What is the total amount the three friends have?

Show hint (soft nudge)
The total never changes — you only need to find it at one moment.
Show hint (sharpest)
Toy's money doubles in the first two rounds, so track it up to just before his own turn.
Show solution
  1. Toy's $36 doubles in each of the first two rounds: 36 → 72 → 144 just before his turn.
  2. He ends with $36, so he gave away 144 − 36 = 108, which exactly doubled Amy and Jan — meaning they held $108 then.
  3. The total, unchanged throughout, is 144 + 108 = $252.
D $252.
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Problem 21 · 1990 AJHSME Stretch
Algebra & Patterns work-backward

A list of 8 numbers is formed by beginning with two given numbers. Each new number in the list is the product of the two previous numbers. Find the first number if the last three numbers are 16, 64, 1024.

Show hint (soft nudge)
Each term is the product of the two before it, so divide to step backward.
Show hint (sharpest)
From 16, 64, 1024 work back: the term before 16 is 64 ÷ 16, and so on.
Show solution
  1. Since 1024 = 16 · 64, the term before 16 is 64 ÷ 16 = 4, then 16 ÷ 4 = 4, then 4 ÷ 4 = 1, then 4 ÷ 1 = 4, and finally 1 ÷ 4 = 1/4.
  2. So the first number is 1/4.
B 1/4.
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Problem 24 · 1990 AJHSME Stretch
Algebra & Patterns substitution balance
ajhsme-1990-24
Show hint (soft nudge)
Write the two balances as equations: 3 triangles + 1 diamond = 9 circles, and 1 triangle = 1 diamond + 1 circle.
Show hint (sharpest)
Substitute to eliminate the triangles.
Show solution
  1. From 3T + D = 9C and T = D + C: 3(D + C) + D = 9C gives 4D = 6C, so 2D = 3C.
  2. Thus two diamonds balance 3 circles.
C 3.
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Problem 21 · 1988 AJHSME Stretch
Algebra & Patterns case-on-where-n-falls

A fifth number, n, is added to the set {3, 6, 9, 10} to make the mean of the set of five numbers equal to its median. The number of possible values of n is

Show hint (soft nudge)
The median depends on where n falls in the sorted list; split into three cases.
Show hint (sharpest)
In each case, set mean = median and solve for n.
Show solution
  1. Mean = (28 + n)⁄5. Case n ≤ 6: median = 6 → n = 2. Case 6 ≤ n ≤ 9: median = n → 28 + n = 5n → n = 7. Case n ≥ 9: median = 9 → n = 17.
  2. Each value of n is consistent with its case, so the answers are 2, 7, 17 — 3 values.
C 3.
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Problem 24 · 1987 AJHSME Stretch
Algebra & Patterns score-constraint parity

A multiple choice examination consists of 20 questions. The scoring is +5 for each correct answer, −2 for each incorrect answer, and 0 for each unanswered question. John's score on the examination is 48. What is the maximum number of questions he could have answered correctly?

Show hint (soft nudge)
Let c = correct, w = wrong; 5c − 2w = 48 with c + w ≤ 20 and c, w ≥ 0 integers.
Show hint (sharpest)
5c must be even, so c must be even — that limits c to even values.
Show solution
  1. From 5c − 2w = 48, c must be even (so 5c is even); from c + w ≤ 20, c is at most about 12.6.
  2. The biggest even c that fits is c = 12 (w = 6, total 18).
D 12.
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Problem 24 · 1985 AJHSME Stretch
Algebra & Patterns double-count-vertices
ajhsme-1985-24
Show hint (soft nudge)
Add the three side-sums: each vertex is counted twice, each midpoint once.
Show hint (sharpest)
3S = 2·(vertex sum) + (midpoint sum) = (vertex sum) + 75.
Show solution
  1. Total of all 6 numbers 10 + 11 + ⋯ + 15 = 75. Adding the three side-sums gives each vertex twice and each midpoint once, so 3S = (vertex sum) + 75. Maximize by putting the three biggest at the vertices: 13 + 14 + 15 = 42.
  2. 3S = 42 + 75 = 117 → S = 39 (with midpoints 12, 10, 11 between the matching vertex pairs).
D 39.
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