AMC 10 · 2022 · #8

Grade 5 arithmetic

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

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Problem

Consider the following $100$ sets of $10$ elements each:
\begin{align*} &{1,2,3,\ldots,10}, \ &{11,12,13,\ldots,20},\ &{21,22,23,\ldots,30},\ &\vdots\ &{991,992,993,\ldots,1000}. \end{align*}
How many of these sets contain exactly two multiples of $7$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

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

Givens: $100$ blocks, each $10$ consecutive integers, covering $1$ through $1000$; Want: count of blocks containing exactly $2$ multiples of $7$; Answer choices: (A) $40$, (B) $42$, (C) $43$, (D) $49$, (E) $50$

Unknowns: The number of blocks with exactly two multiples of $7$

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #9 Solve an Easier Related Problem, #15 Organize Information in More Ways

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.

Execute — Answer: B

#9 Solve an Easier Related Problem 4.OA.B.4 Step 1

Try a small case first to feel the structure.
Block $\{1, \ldots, 10\}$ contains $7$ — one multiple of $7$.
Block $\{11, \ldots, 20\}$ contains $14$ — one multiple.
Block $\{21, \ldots, 30\}$ contains $21$ and $28$ — two multiples.
So in the first three blocks: $1, 1, 2$.

$$\text{first three blocks contain}\;1,\;1,\;2\;\text{multiples of}\;7$$

💡 Looking at three small blocks shows that each block always has either one or two multiples — never zero.

#15 Organize Information in More Ways 4.OA.B.4 Step 2

Argue every block has at least one multiple of $7$.
Consecutive multiples of $7$ are $7$ apart, so in any $10$ consecutive integers there is at least one multiple of $7$ (the gap is shorter than the block width).
Argue no block has three: three consecutive multiples of $7$ span $14$, which is bigger than $9$, the width of a block from first to last index.
So every block contains either $1$ or $2$ multiples of $7$.

$$\text{gap}=7 < 10,\;\text{so}\ge 1;\;\;14 > 9,\;\text{so}\le 2$$

💡 Spacing of $7$ inside a window of $10$ forces exactly one or exactly two — counting by spacing instead of by block.

#16 Change Focus / Count the Complement 5.NBT.B.6 Step 3

Count the total number of multiples of $7$ from $1$ to $1000$.
The largest multiple of $7$ within $1000$ is $7 \times 142 = 994$ (since $7 \times 143 = 1001 > 1000$).
So there are $142$ multiples of $7$ between $1$ and $1000$ inclusive.

$$\lfloor 1000 \div 7 \rfloor = 142,\;\text{since}\;7 \times 142 = 994$$

💡 Count the multiples directly — that is the 'flipped' question.

#15 Organize Information in More Ways 4.OA.A.3 Step 4

Combine.
Let $b$ be the number of blocks with two multiples of $7$, and $a$ the number with exactly one.
Then $a + b = 100$ (every block falls in exactly one category) and total multiples = $1 \cdot a + 2 \cdot b = 142$ (sum of per-block counts).
Subtract: $(a + 2b) - (a + b) = 142 - 100$, giving $b = 42$.

$$a+b = 100,\;\;a + 2b = 142 \;\Rightarrow\; b = 42$$

💡 Pretend every block had one multiple — that accounts for $100$. The extra $42$ multiples are the 'surplus', one per block that has two.

#16 Change Focus / Count the Complement 4.NBT.B.4 Step 5

Answer: $42$ blocks contain exactly two multiples of $7$.
That is choice (B).

$$b = 42 \;\Rightarrow\; \textbf{(B)}$$

💡 Reading the surplus directly as the answer — the change of focus pays off.

[1] #9 4.OA.B.4 Try a small case first to feel the structure. Block $\{1, \ldots, 10\}$ contains

[2] #15 4.OA.B.4 Argue every block has at least one multiple of $7$. Consecutive multiples of $7$

[3] #16 5.NBT.B.6 Count the total number of multiples of $7$ from $1$ to $1000$. The largest multi

[4] #15 4.OA.A.3 Combine. Let $b$ be the number of blocks with two multiples of $7$, and $a$ the

[5] #16 4.NBT.B.4 Answer: $42$ blocks contain exactly two multiples of $7$. That is choice (B).

Review

Reasonableness: Sanity check on extremes: total multiples of $7$ in $1$–$1000$ is $\lfloor 1000/7 \rfloor = 142$; 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, $7 \times 10$), the number of multiples is $\lfloor 70/7 \rfloor = 10$, distributed roughly as $3$ blocks-with-$2$ and $4$ blocks-with-$1$ ($3 \cdot 2 + 4 \cdot 1 = 10$). Scaled to $100$ blocks: about $\tfrac{3}{7} \cdot 100 \approx 43$ 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.

Alternative: Tool #2 (Systematic List): list multiples of $7$ explicitly — $7, 14, 21, 28, 35, 42, 49, 56, \ldots$ — and tag each with the block it lands in. A block holds two multiples exactly when there are two values of $k$ with $\lceil 7k / 10 \rceil$ equal. Counting these directly gives the same $42$. This is slower than the complement count of multiples, but it is the brute-force verification path.

CCSS standards used (min grade 5)

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Identifying which numbers are multiples of $7$ inside each block and reasoning about their spacing.)
5.NBT.B.6 Find whole-number quotients with up to four-digit dividends and two-digit divisors (Computing $\lfloor 1000 / 7 \rfloor = 142$ to count all multiples of $7$ up to $1000$.)
4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Combining 'each block has 1 or 2 multiples' with 'total is 142' to deduce $b = 142 - 100 = 42$.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Final subtraction $142 - 100 = 42$ to read off the answer.)

⭐ 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 5)

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