AMC 8 · Test Mode

2026 AMC 8

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Problem 1 · 2026 AMC 8 Easy
Arithmetic & Operations grouping arithmetic-series

What is the value of the following expression?

1 + 2 − 3 + 4 + 5 − 6 + 7 + 8 − 9 + 10 + 11 − 12
Show hint (soft nudge)
Doing it left-to-right is slow. Look at the sign pattern (+, +, −) and see if it suggests a group size.
Show hint (sharpest)
The signs repeat +, +, −. Group the terms in threes and watch the group totals.
Show solution
  1. Group in threes: (1+2−3), (4+5−6), (7+8−9), (10+11−12).
  2. The totals climb by 3 each time: 0, 3, 6, 9.
  3. Sum: 0 + 3 + 6 + 9 = 18.
A The answer is 18.
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Problem 2 · 2026 AMC 8 Easy
Arithmetic & Operations symmetry careful-counting

In the array shown below, three 3s are surrounded by 2s, which are in turn surrounded by a border of 1s. What is the sum of the numbers in the array?

1111111
1222221
1233321
1222221
1111111
A 5 × 7 array of numbers.
Show hint (soft nudge)
Several rows of the grid are duplicates. Add one of each kind, then count copies.
Show hint (sharpest)
Use the symmetry: the top and bottom rows match, and the 2nd and 4th match — so you only really add three different rows.
Show solution
  1. Top and bottom rows are all 1s: 7 + 7 = 14.
  2. Second and fourth rows: 1 + (2+2+2+2+2) + 1 = 12 each, so 12 + 12 = 24.
  3. Middle row: 1 + 2 + 3 + 3 + 3 + 2 + 1 = 15.
  4. Total: 14 + 24 + 15 = 53.
C The answer is 53.
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Problem 3 · 2026 AMC 8 Medium
Geometry & Measurement perimeter pythagorean-triple square-area

Haruki has a piece of wire that is 24 centimeters long. He wants to bend it to form each of the following shapes, one at a time.

  • A regular hexagon with side length 5 cm.
  • A square of area 36 cm2.
  • A right triangle whose legs are 6 and 8 cm long.

Which of the shapes can Haruki make?

Show hint (soft nudge)
The wire is 24 cm and won't stretch. For each shape, just ask: would its perimeter be exactly 24?
Show hint (sharpest)
The wire can't stretch, so a shape works only if its perimeter is exactly 24 cm. Find each perimeter.
Show solution
  1. Hexagon: perimeter = 6 × 5 = 30 cm. Longer than 24 — not possible.
  2. Square of area 36: side = √36 = 6, so perimeter = 4 × 6 = 24 cm. Possible. ✓
  3. Right triangle with legs 6 and 8: hypotenuse = √(62+82) = √100 = 10, so perimeter = 6 + 8 + 10 = 24 cm. Possible. ✓
  4. Only the square and the triangle can be made.
D Square and triangle only.
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Problem 4 · 2026 AMC 8 Medium
Fractions, Decimals & Percents percent-multiplier

Brynn's savings decreased by 20% in July, then increased by 50% of the new amount in August. Brynn's savings are now what percent of the original amount?

Show hint (soft nudge)
Don't pick a starting amount — that's extra work. Each percent change is something you can just multiply by.
Show hint (sharpest)
Each percent change is just a multiplier — you don't even need a starting amount. Multiply the two.
Show solution
  1. Down 20% means × 0.8; up 50% means × 1.5.
  2. Multiply the changes: 0.8 × 1.5 = 1.2.
  3. 1.2 = 120% of the original.
E 120%.
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Problem 5 · 2026 AMC 8 Stretch
Ratios, Rates & Proportions distance-speed-time

Casey went on a road trip that covered 100 miles, stopping only for a lunch break along the way. The trip took 3 hours in total and her average speed while driving was 40 miles per hour. In minutes, how long was the lunch break?

Show hint (soft nudge)
The 3 hours includes the break. Find the driving time first.
Show hint (sharpest)
Time = distance ÷ speed gives only the driving time. Whatever's left of the 3 hours is the break.
Show solution
  1. Driving time = 100 ÷ 40 = 2.5 hours.
  2. Break = total − driving = 3 − 2.5 = 0.5 hour.
  3. 0.5 hour = 30 minutes.
B 30 minutes.
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Problem 6 · 2026 AMC 8 Medium
Geometry & Measurement area-decomposition
amc8-2026-06
Show hint (soft nudge)
The part Peter can't reach is the inner rectangle more than 1 m from every edge.
Show hint (sharpest)
Shrink each dimension by 2 (one meter on each side).
Show solution
  1. The unreachable middle is (10 − 2) × (8 − 2) = 48, out of the field's 10 × 8 = 80.
  2. So the reachable border is 80 − 48 = 32, a fraction 32/80 = 2/5.
E 2/5.
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Problem 7 · 2026 AMC 8 Medium
Fractions, Decimals & Percents fraction-of

Mika wants to estimate how far a new electric bike goes on a full charge. She made two trips totaling 40 miles: the first used 12 of the battery and the second used 310 of the battery. How many miles can the bike go on a fully charged battery?

Show hint (soft nudge)
Add the two battery fractions to see what share of a full charge covered the 40 miles.
Show hint (sharpest)
Then scale up from that share to a whole battery.
Show solution
  1. The two trips used ½ + 3/10 = 4/5 of the battery for 40 miles.
  2. A full battery covers 40 ÷ (4/5) = 50 miles.
C 50 miles.
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Problem 8 · 2026 AMC 8 Medium
Fractions, Decimals & Percents reduce-fraction

A poll asked some people whether they liked solving mathematics problems, and exactly 74% answered "yes." What is the fewest possible number of people who could have been asked?

Show hint (soft nudge)
74% of the group must be a whole number of people.
Show hint (sharpest)
Reduce 74/100 to lowest terms; the denominator is the smallest possible group size.
Show solution
  1. 74% = 74/100 = 37/50 in lowest terms, so the number of people must be a multiple of 50.
  2. The fewest is 50.
D 50 people.
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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 10 · 2026 AMC 8 Medium
Logic & Word Problems ordering

Five runners finished a race: Luke, Melina, Nico, Olympia, and Pedro. Nico finished 11 minutes behind Pedro. Olympia finished 2 minutes ahead of Melina but 3 minutes behind Pedro. Olympia finished 6 minutes ahead of Luke. Which runner finished fourth?

Show hint (soft nudge)
Place everyone on a timeline measured from Pedro.
Show hint (sharpest)
Then read off who is fourth from the front.
Show solution
  1. Measuring minutes behind Pedro: Olympia +3, Melina +5 (2 behind Olympia), Luke +9 (6 behind Olympia), Nico +11.
  2. The order is Pedro, Olympia, Melina, Luke, Nico, so fourth is Luke.
A Luke.
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Problem 11 · 2026 AMC 8 Medium
Geometry & Measurement arc-length
amc8-2026-11
Show hint (soft nudge)
A quarter circle in a square of side s has arc length one-fourth of 2πs.
Show hint (sharpest)
Add up the arcs for all five squares.
Show solution
  1. Each square of side s contributes a quarter-circle of length ¼ · 2πs = πs/2.
  2. Total = (π/2)(1 + 1 + 2 + 3 + 5) = (π/2)(12) = .
B 6π.
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Problem 12 · 2026 AMC 8 Hard
Logic & Word Problems constraint-propagation
amc8-2026-12
Show hint (soft nudge)
Start at the most restrictive sum and see which digits can make it.
Show hint (sharpest)
A sum of 10 between two circles can only be 4 and 6.
Show solution
  1. The edge summing to 10 forces those two circles to be 4 and 6; taking the upper one as 4 makes the top circle 9 − 4 = 5 and the bottom-left 6.
  2. Then the remaining sums give 8 − 6 = 2, 5 − 2 = 3, 4 − 3 = 1, and 5 + 1 = 6 checks out — using each digit 1–6 once.
  3. So the top circle must be 5.
D 5.
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Problem 13 · 2026 AMC 8 Hard
Geometry & Measurement tilted-square pythagorean
amc8-2026-13
Show hint (soft nudge)
The square is tilted, so its area equals the square of its side length.
Show hint (sharpest)
Read one side as a step across the lattice and use the Pythagorean theorem.
Show solution
  1. Each side of the shaded square is the hypotenuse of a right triangle with legs 3 and 1, measured across the tiling.
  2. So the area is the side squared: 3² + 1² = 10.
A 10.
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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 15 · 2026 AMC 8 Hard
Geometry & Measurement spatial-reasoning
amc8-2026-15
Show hint (soft nudge)
A cube's two shaded faces share an edge, so both must be glued to neighbors to hide them.
Show hint (sharpest)
Look for an arrangement where every cube has two glued faces that meet at an edge.
Show solution
  1. To hide a cube's two shaded faces — which meet at an edge — it must be glued to neighbors on two faces sharing an edge.
  2. Four cubes arranged in a 2 × 2 square give each cube exactly two such adjacent glued faces; with three or fewer, some cube has only one glued face (or two opposite ones), so a shaded face shows.
  3. The fewest is 4.
A 4 cubes.
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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 17 · 2026 AMC 8 Hard
Counting & Probability casework

Four students sit in a row and chat with the people next to them. They then rearrange themselves so that no one is seated next to anyone they sat next to before. How many such rearrangements are possible?

Show hint (soft nudge)
Label the seats 1, 2, 3, 4; the forbidden neighbor-pairs are 1-2, 2-3, and 3-4.
Show hint (sharpest)
Try to build a valid order — the options turn out to be very tight.
Show solution
  1. No two originally-adjacent students may be neighbors, so none of 1-2, 2-3, 3-4 can touch.
  2. The only orders that work are 2-4-1-3 and its reverse 3-1-4-2, giving 2 rearrangements.
A 2.
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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 19 · 2026 AMC 8 Hard
Ratios, Rates & Proportions relative-distance

Miguel and his dog Luna start together at a park entrance. Miguel throws a ball straight ahead to a tree and keeps walking at a steady pace. Luna sprints to the ball and immediately brings it back to Miguel. Luna runs 5 times as fast as Miguel walks. What fraction of the entrance-to-tree distance does Miguel cover by the time Luna brings him the ball?

Show hint (soft nudge)
In the same time, Luna covers 5 times the distance Miguel does.
Show hint (sharpest)
Luna's path is entrance → tree → Miguel; set its length to 5 times Miguel's walk.
Show solution
  1. Let the entrance-to-tree distance be 1 and Miguel's distance be d. Luna runs to the tree and back to Miguel: 1 + (1 − d) = 2 − d.
  2. Since Luna covers 5 times Miguel's distance, 5d = 2 − d, so 6d = 2 and d = 1/3.
D 1/3.
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Problem 20 · 2026 AMC 8 Stretch
Counting & Probability recursion compositions

The land of Catania uses gold coins (1 mm thick) and silver coins (3 mm thick). In how many ways can Taylor make a stack exactly 8 mm tall using any arrangement of gold and silver coins, where order matters?

Show hint (soft nudge)
Let f(n) count stacks of height n; the top coin is either gold (leaving n − 1) or silver (leaving n − 3).
Show hint (sharpest)
Build the counts up from small heights using f(n) = f(n−1) + f(n−3).
Show solution
  1. With f(0) = f(1) = f(2) = 1, the rule f(n) = f(n−1) + f(n−3) gives 1, 1, 1, 2, 3, 4, 6, 9, 13.
  2. So f(8) = 13 stacks.
D 13.
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Problem 21 · 2026 AMC 8 Stretch
Counting & Probability markov-chain casework
amc8-2026-21
Show hint (soft nudge)
Each outer point connects only to inner points; each inner point connects to 2 outer and 2 inner points.
Show hint (sharpest)
Track the chance of being on an outer point after each move.
Show solution
  1. From an outer point both edges lead inward, so after move 1 the spider is surely on an inner point.
  2. From an inner point it returns to an outer point with probability ½, so after move 2 it is outer with probability ½ (inner with probability ½).
  3. To end outer after move 3 it must be inner after move 2 (½) and then step outward (½): ½ · ½ = 1/4.
B 1/4.
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Problem 22 · 2026 AMC 8 Stretch
Logic & Word Problems extremal median

The integers 1 through 25 are arbitrarily separated into five groups of 5 numbers each. The median of each group is found, and M is the median of those five medians. What is the least possible value of M?

Show hint (soft nudge)
M is the 3rd-smallest of the five medians, so you want three groups with small medians.
Show hint (sharpest)
Each small median needs two even-smaller numbers beneath it, and small numbers are in limited supply.
Show solution
  1. To make M small, three groups need small medians, each requiring two smaller numbers below it. Using 1–6 as those "below" numbers lets three groups have medians 7, 8, and 9, while the other two groups absorb the large numbers.
  2. Trying for M = 8 fails: three medians ≤ 8 would need six distinct numbers below them all drawn from {1,…,5}, which isn't enough.
  3. So the least possible value is 9.
A 9.
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Problem 23 · 2026 AMC 8 Stretch
Geometry & Measurement belt-around-circles
amc8-2026-23
Show hint (soft nudge)
The curved pieces of a tight band around a convex bunch always add up to one full circle.
Show hint (sharpest)
The straight pieces equal the perimeter of the shape joining the outermost centers.
Show solution
  1. The band turns 360° all the way around, so its curved pieces total one full circle: 2π · 2 = 4π.
  2. The straight pieces equal the perimeter of the trapezoid joining the four outer coin centers (radius-2 coins touching): 8 + 4 + 4 + 4 = 20.
  3. So the band is 4π + 20 centimeters.
C 4π + 20.
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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 25 · 2026 AMC 8 Stretch
Counting & Probability corner-cutting casework
amc8-2026-25
Show hint (soft nudge)
Such a hexagon is the equilateral triangle with three corner equilateral triangles cut off.
Show hint (sharpest)
If the triangle has side 6, the three integer cut sizes must leave each middle segment ≥ 1, so each pair of cuts sums to at most 5.
Show solution
  1. The triangle ABC has side 6 (from the example: 1 + 3 + 2). Cutting integer-sized corners a, b, c leaves middle segments 6 − a − b, 6 − b − c, 6 − c − a, which must each be ≥ 1, so every pair of cuts sums to at most 5.
  2. Counting unordered cut-triples {a, b, c} of positive integers with all pairwise sums ≤ 5 gives {1,1,1}, {1,1,2}, {1,2,2}, {2,2,2}, {1,1,3}, {1,2,3}, {2,2,3}, {1,1,4} — that's 8 hexagons (rotations and reflections counted once).
E 8.
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