AMC 10 · 2022 · #8

Grade 5 arithmetic

Archetype Modular Cycle Frequency Count └ Sub-archetype Floor Count Across Residue Bins

multiplesmodular-arithmeticdivisibility-rulespattern-recognitioncomplementary-counting complementary-countingpattern-recognitioneasier-related-problem ↑ Prerequisites: multiplesdivisibility-rules

📏 Medium solution 💡 2 insights

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Problem

The integers from 1 to 1000 are split into 100 blocks of 10 consecutive numbers each: {1, 2, …, 10}, {11, …, 20}, …, {991, …, 1000}. How many of these blocks contain exactly two multiples of 7?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

Counting blocks one by one for 100 blocks is grinding. Tool #16 (Change Focus) flips the question: instead of counting blocks, count the multiples of 7 from 1 to 1000 and notice how they distribute across blocks. Tool #15 (Reorganize) groups multiples by their host block — each block has either 1 or 2 multiples (never 0 or 3+ in a window of 10). Tool #9 (Easier Problem) anchors why: spacing between multiples of 7 is 7, and a block of width 10 can fit at most two. So if we know the total count of multiples and the minimum per block, the leftover tells us how many blocks have an extra one. No algebra needed.

1STEP 1

Test small blocks: {1–10} holds one multiple of 7, {11–20} one, {21–30} two (21 and 28) — counts run 1, 1, 2.

first three blocks contain 1, 1, 2 multiples of 7

2STEP 2

Multiples of 7 sit 7 apart; three would span 14 ＞ 9 (a block's width), so every block holds 1 or 2 multiples.

gap=7 ＜ 10, so ≥ 1; 14 ＞ 9, so ≤ 2

3STEP 3

Count all multiples of 7 up to 1000: the largest is 7 × 142 = 994 (7 × 143 = 1001 ＞ 1000), so there are 142 of them.

⌊ 1000 ÷ 7 ⌋ = 142, since 7 × 142 = 994

4STEP 4

With a + b = 100 blocks and a + 2b = 142 multiples, subtracting the equations gives b = 42.

a+b = 100, a + 2b = 142 → b = 42

5STEP 5

So 42 blocks contain exactly two multiples of 7 — choice (B).

b = 42 → (B)

Answer

Sanity check on extremes: total multiples of 7 in 1–1000 is ⌊

\frac{1000}{7}

⌋ = 142; smallest count per block is 1, largest is 2. The answer 42 sits between 0 and 100 as required. Spot-check the structure: in any 7 consecutive blocks (70 numbers, 7 × 10), the number of multiples is ⌊

\frac{70}{7}

⌋ = 10, distributed roughly as 3 blocks-with-2 and 4 blocks-with-1 (3 · 2 + 4 · 1 = 10). Scaled to 100 blocks: about

\frac{3}{7}

· 100 ≈ 43 blocks-with-2, which is the rough density. The exact answer 42 matches choice (B); the close miss 43 in (C) would come from the rough average — the precise total 142 favors 42 over 43. Choices 40, 49, 50 are too far from the density argument.

💡Key takeaway

This AMC 10 problem only needs Grade 5 division and the multiples idea you already know — there are 142 multiples of 7 up to 1000 spread across 100 blocks, every block has at least one, so the extra 142 - 100 = 42 tells you exactly how many blocks have two.

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