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AMC 10 · 2022 · #8

Easy mode Grade 5
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Problem

Picture all the whole numbers from 11 to 10001000, split into groups of 1010 in a row:

{1,2,3,…,10}\{1, 2, 3, \ldots, 10\}, then {11,12,13,…,20}\{11, 12, 13, \ldots, 20\}, then {21,22,23,…,30}\{21, 22, 23, \ldots, 30\}, and so on, all the way up to {991,992,993,…,1000}\{991, 992, 993, \ldots, 1000\}.

That gives us 100100 groups in total. Each group has 1010 numbers in a row.

Now look at each group and count the multiples of 77 inside it. We want the groups that have exactly two multiples of 77.

How many of the 100100 groups have exactly two multiples of 77?

Pick an answer.

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

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

#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.

⭐ 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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