# Difference between revisions of "2011 AMC 12A Problems/Problem 7"

## Problem

A majority of the $30$ students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than $1$. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $17.71$. What was the cost of a pencil in cents?

$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 11 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 77$

## Solution

The total cost of the pencils can be found by $(\text{students}*\text{pencils purchased by each}*\text{price of each pencil})$. As the cost is $17.71$ dollars, or $1771$ cents, the cost of the pencils must divide the total cost. Scanning the answer choices, they all unfortunately divide into $1771$, so we have to start from the beginning (although this does give us some indication that we are on the right track, because MAA doesn't want to make things too easy for us).

Therefore, since $1771$ is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: $7, 11, 23$. Since neither $(C)$ nor $(E)$ are any of these factors, they can be eliminated immediately, leaving $(A)$, $(B)$, and $(D)$.

Beginning with $(A) 7$, we see that the number of pencils purchased by each student must be either $11$ or $23$. However, the problem states that the price of each pencil must exceed the number of pencils purchased, so we can eliminate this.

Continuing with $(B) 11$, we can conclude that the only case that fulfills the restrictions are that there are $23$ students who each purchased $7$ such pencils, so the answer is $\boxed{B}$. We can apply the same logic to $(E)$ as we applied to $(A)$, if one wants to make doubly sure.

 2011 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions