AMC 10 · 2022 · #8
Grade 5 arithmeticProblem
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.
AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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) $50$
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
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: $1, 1, 2$.
💡 Looking at three small blocks shows that each block always has either one or two multiples — never zero.
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 $7$.
💡 Spacing of $7$ inside a window of $10$ forces exactly one or exactly two — counting by spacing instead of by block.
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.
💡 Count the multiples directly — that is the 'flipped' question.
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 $b = 42$.
💡 Pretend every block had one multiple — that accounts for $100$. The extra $42$ multiples are the 'surplus', one per block that has two.
4.NBT.B.4 Step 5 - Answer: $42$ blocks contain exactly two multiples of $7$.
- That is choice (B).
💡 Reading the surplus directly as the answer — the change of focus pays off.
4.OA.B.4 Try a small case first to feel the structure. Block $\{1, \ldots, 10\}$ contains 4.OA.B.4 Argue every block has at least one multiple of $7$. Consecutive multiples of $7$ 5.NBT.B.6 Count the total number of multiples of $7$ from $1$ to $1000$. The largest multi 4.OA.A.3 Combine. Let $b$ be the number of blocks with two multiples of $7$, and $a$ the 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.4Find 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.6Find 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.3Solve 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.4Fluently 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.
⭐ 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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