Competition · AMC preparation · step 4 of 4

AMC 10 · 2022B · #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
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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)
40
(B)
42
(C)
43
(D)
49
(E)
50

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

Try the first few blocks

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

Bound each block's count

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 multiples of 7

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

Solve the two equations

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

Read off the block count

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

b = 42 → (B)
Answer
42
Sanity check on extremes: total multiples of 7 in 1–1000 is ⌊ 10007\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 ⌊ 707\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 37\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.

  • Try the first few blocks
  • Bound each block's count
  • Count multiples of 7
  • Solve the two equations
  • Read off the block count

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