2011 AMC 10A Problem 10

Let $s$ be the number of students who bought pencils, $n$ the number of pencils each bought, and $c$ the cost of a pencil in cents. Then $s \cdot n \cdot c = 1771$, with $16 \le s \le 30$ (a majority of $30$), $n > 1$, and $c > n$.

Factor $1771$: it is divisible by $7$, giving $1771 = 7 \cdot 253$, and $253 = 11 \cdot 23$. So
$$1771 = 7 \cdot 11 \cdot 23.$$

The only divisor of $1771$ between $16$ and $30$ is $23$ (the others are $1, 7, 11, 77, 161, 253, 1771$). Hence $s = 23$ and $n \cdot c = 77 = 7 \cdot 11$. With $1 < n < c$, we need $n = 7$ and $c = 11$.

The answer is $\boxed{\textbf{(B)}\ 11}$.

When a product of unknown integers is given, factor it into primes and let the constraints decide which factor goes where. Here the bound on the number of students identified one factor immediately, and the ordering condition $c > n$ sorted the remaining two. Always translate word constraints ("majority," "more than 1") into inequalities before assigning factors.

• Failing to factor 1771 (it is not prime) or ignoring the 'majority' condition, which is what pins the student count to 23.
• Answering $7$ by swapping the roles of pencils per student and price per pencil, which violates the condition that the price exceeds the count.