2022 AMC 10B Problem 8

Ten consecutive integers always include a multiple of $7$ (any seven consecutive integers do), and never three of them (that would need a span of at least $14$). So every set contains exactly one or exactly two multiples of $7$.

The total number of multiples of $7$ from $1$ to $1000$ is $\left\lfloor \frac{1000}{7} \right\rfloor = 142$.

Let $x$ sets contain two multiples and $100 - x$ contain one. Counting multiples of $7$ two ways,
$$2x + (100 - x) = 142 \quad\Longrightarrow\quad x = 42.$$

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

Instead of inspecting each block, bound what a block can contain (here, $1$ or $2$) and then use a global total. This "count the same thing two ways" step turns a messy positional question into one linear equation. The bound is the whole trick: once each block has one or two, the total pins down how many have two.

Multiples of $7$ repeat every $70$ numbers, which is $7$ blocks. In $1$–$70$ the multiples are $7, 14, 21, 28, 35, 42, 49, 56, 63, 70$; blocks $3$ ($21, 28$), $5$ ($42, 49$) and $7$ ($63, 70$) hold two, so exactly $3$ of every $7$ blocks contain two multiples. Fourteen full cycles cover $980$ numbers ($98$ blocks) and give $42$; the last two blocks ($981$–$1000$) contain $987$ and $994$, one each. Total $42$.

• Trying to track the position of the multiples of 7 block by block, or assuming every block has at least one multiple of 7 only after checking a few.
• Computing $142 - 100 = 42$ correctly but then "adjusting" for the final block, or answering the number of one-multiple blocks ($58$).