2022 AMC 10B Problem 8

Checking $100$ sets one by one is too slow. Instead, first work out how many multiples of $7$ a single set can hold, then use a total.

1. At least one. Among any $7$ numbers in a row, one is a multiple of $7$. Each set has $10$ numbers in a row, so it has at least one.
2. At most two. Three multiples of $7$ in a row, like $7, 14, 21$, stretch from the first to the last over $14$ steps. A set only stretches $9$ steps (from $1$ to $10$, say). So a set never holds three.
3. Examples. $\{1, \ldots, 10\}$ holds only $7$. $\{21, \ldots, 30\}$ holds $21$ and $28$. So every set holds exactly one or exactly two.
4. The total. From $1$ to $1000$ there are $142$ multiples of $7$, because $7 \times 142 = 994$ and $7 \times 143 = 1001$ is too big.
5. Split the total. Give each of the $100$ sets one multiple first. That uses $100$ of them. The $142 - 100 = 42$ left over must go one each into $42$ different sets (a set cannot take a third). So $42$ sets have two.

The answer is $\boxed{\textbf{(B)}\ 42}$.

When you cannot easily track each group, find the smallest and largest amount one group can hold, then use the grand total. Here every group holds one or two, so the extra multiples beyond one per group tell you exactly how many groups hold two. This "count the same things two ways" trick turns a long check into one subtraction.

Look for the repeating pattern. Multiples of $7$ repeat every $70$ numbers, which is $7$ sets. In $1$ to $70$ the multiples are $7, 14, 21, 28, 35, 42, 49, 56, 63, 70$. The sets $21$–$30$ ($21, 28$), $41$–$50$ ($42, 49$) and $61$–$70$ ($63, 70$) hold two, so $3$ of every $7$ sets hold two. Fourteen full rounds cover $1$ to $980$ ($98$ sets) and give $14 \times 3 = 42$. The last two sets, $981$–$990$ and $991$–$1000$, hold only $987$ and $994$, one each. Total: $42$.