AMC 10 · 2022 · #6

Grade 5 arithmetic
multiplesmodular-arithmeticdivisibility-rulespattern-recognitioncomplementary-counting complementary-countingpattern-recognitioneasier-related-problem ↑ Prerequisites: multiplesdivisibility-rules
📏 Medium solution 💡 2 insights
📘 View easy version →
Problem
The integers from 1 to 1000 are split into 100 blocks of ten consecutive numbers each. Find how many of these blocks contain exactly two multiples of seven.

Pick an answer.

(A)
40
(B)
42
(C)
43
(D)
49
(E)
50
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

Look at the first few blocks

The first blocks hold one or two.

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

Bound one block

No block can hold three.

gap=7 < 10, so ≥ 1; 14 > 9, so ≤ 2
3STEP 3

Count all the multiples

Count all the multiples of seven.

⌊ 1000 ÷ 7 ⌋ = 142, since 7 × 142 = 994
4STEP 4

Set up two equations

Block count and multiple count give two equations.

a+b = 100, a + 2b = 142 → b = 42
5STEP 5

Solve and read the answer

Solving gives 42.

b = 42 → (B)
Answer
42
Sanity check on extremes: total multiples of 7 in 1–1000 is ⌊ 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 ⌊ 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 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 12 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.