2021 AMC 10A Problem 22

Suppose the roommate took $c$ consecutive sheets starting right after sheet $b$, so the missing pages are $2b+1, 2b+2, \dots, 2b+2c$: that is $2c$ pages with average $\frac{(2b+1) + (2b+2c)}{2}$, hence total $c(4b + 2c + 1)$.

All $50$ pages sum to $\frac{50 \cdot 51}{2} = 1275$. The $50 - 2c$ remaining pages have mean $19$, so their total is $19(50 - 2c) = 950 - 38c$. Therefore
$$1275 - c(4b + 2c + 1) = 950 - 38c \quad\Longrightarrow\quad c(4b + 2c - 37) = 325 = 5^2 \cdot 13.$$

Now $c$ must be a divisor of $325$ with $1 \le c < 25$: $c \in \{1, 5, 13\}$.
- $c = 1$: $4b - 35 = 325$, $b = 90$, impossible.
- $c = 5$: $4b - 27 = 65$, $b = 23$, but then sheets $24$ to $28$ would be taken; only $25$ exist.
- $c = 13$: $4b - 11 = 25$, $b = 9$: sheets $10$ through $22$, pages $19$ to $44$. Valid.

Check: the remaining $24$ pages ($1$ to $18$ and $45$ to $50$) total $171 + 285 = 456 = 19 \cdot 24$.

The answer is $\boxed{\textbf{(B)}\ 13}$.

A mean condition is a linear equation in totals, and totals of consecutive integers are easy. Because the unknowns are integers, the resulting equation is Diophantine: rearranging to (unknown) $\times$ (expression) $=$ constant makes the divisor structure of $325$ do the searching for you, and the physical limits ($25$ sheets) discard the spurious factorizations.

Test the choices. Removing $c$ sheets leaves $50 - 2c$ pages totaling $950 - 38c$, so the removed pages total $325 + 38c$, which must equal $c$ times an odd number $4b + 2c + 1$. For $c = 13$: $325 + 494 = 819 = 13 \cdot 63$, and $4b + 27 = 63$ gives $b = 9$. The other choices fail: e.g. $c = 10$ gives $705/10$, not an integer.

• Mixing pages and sheets (c sheets removed means 2c pages, and the removed pages run from 2b+1 to 2b+2c).
• Accepting $c = 5$ without checking that the sheets taken must exist ($b + c \le 25$).