AMC 8 · 2025 · #13

Grade 4 number-theory
modular-arithmeticpattern-recognitionsequences-arithmeticgraph-reading pattern-recognitionsystematic-enumerationmodular-arithmetic ↑ Prerequisites: modular-arithmeticmulti-digit-arithmetic
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Problem
Take the 25 even numbers 2, 4, 6, …, 50 and divide each one by 7. Record the remainder (0 through 6) every time. Among the five histograms, find the one whose bar heights match how many times each remainder shows up.

Pick an answer.

(A)
(histogram) bar heights for remainders 0–6: 3, 4, 4, 3, 4, 3, 4
(B)
(histogram) bar heights for remainders 0–6: 3, 4, 4, 4, 3, 3, 4
(C)
(histogram) bar heights for remainders 0–6: 3, 4, 4, 4, 4, 3, 3
(D)
(histogram) bar heights for remainders 0–6: 4, 3, 4, 3, 4, 3, 4
(E)
(histogram) bar heights for remainders 0–6: 4, 4, 3, 4, 3, 4, 3

AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

Dividing 2, 4, 6, … by 7 produces a repeating cycle of remainders, so Tool #5 (Look for a Pattern) is the natural primary tool — compute the first several remainders, spot the cycle, then count efficiently. Tool #2 (Systematic List) supports this by listing the first seven remainders in order so the cycle is visible. Tool #3 (Eliminate Possibilities) is the multiple-choice closer: once we have the seven counts, only one of (A)-(E) matches and the rest can be crossed out.

1STEP 1

The even numbers 2, 4, …, 50 are 2·1 through 2·25, so there are exactly 25 of them — the seven bars must sum to 25.

25 = 2 · 1, 2 · 2, …, 2 · 25 → 25 numbers
2STEP 2

Listing the remainders of the first even numbers when divided by 7 exposes a length-7 cycle: 2, 4, 6, 1, 3, 5, 0.

2 ÷ 7 → 2, 4 ÷ 7 → 4, 6 ÷ 7 → 6, 8 ÷ 7 → 1, 10 ÷ 7 → 3, 12 ÷ 7 → 5, 14 ÷ 7 → 0
3STEP 3

16 leaves remainder 2, restarting the cycle; adding 14 (= 2·7) never changes the remainder mod 7, so the 7-block repeats forever.

16 ÷ 7 → 2 (same as 2 ÷ 7)
4STEP 4

Dividing 25 by 7 gives 25 = 7·3 + 4, so the cycle runs 3 full times (21 numbers) with 4 leftover entries.

25 = 7 · 3 + 4
5STEP 5

After 3 cycles every remainder sits at 3; the leftovers 44, 46, 48, 50 give remainders 2, 4, 6, 1, so bump those four bins by 1.

44 ÷ 7 → 2, 46 ÷ 7 → 4, 48 ÷ 7 → 6, 50 ÷ 7 → 1
6STEP 6

In remainder order 0–6 the bar heights are 3, 4, 4, 3, 4, 3, 4, summing to 25 — matching the even-number count.

[3, 4, 4, 3, 4, 3, 4], 3+4+4+3+4+3+4 = 25 ✓
7STEP 7

Only histogram (A) shows [3, 4, 4, 3, 4, 3, 4]; (B), (C), (D), (E) each differ in a bar, so the answer is (A).

(A) = [3, 4, 4, 3, 4, 3, 4] → (A)
Answer
(histogram) bar heights for remainders 0–6: 3, 4, 4, 3, 4, 3, 4
There are 25 even numbers and 7 possible remainders, so on average each remainder should land 25 / 7 ≈ 3.57 times. With whole-number counts, we'd expect a mix of 3s and 4s — exactly what (A) shows (3 threes and 4 fours). The bar heights sum to 25 as required, and the 4 "extra" remainders (1, 2, 4, 6) are exactly the remainders of the four leftover numbers 50, 44, 46, 48, so the count is internally consistent.
💡Key takeaway

This AMC 8 problem only needs Grade 4 division-with-remainder and pattern-spotting you already know!