2011 AMC 12A Problems/Problem 7

Revision as of 21:44, 14 February 2016 by Elipticmodulo (talk | contribs) (Solution)


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?

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


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 $$ (Error compiling LaTeX. ! Missing $ inserted.)17.71$, 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 tack, because MAA doesn't want to make things too easy for us).

Therefore, since$ (Error compiling LaTeX. ! Missing $ inserted.)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$ (Error compiling LaTeX. ! Missing $ inserted.)(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$ (Error compiling LaTeX. ! Missing $ inserted.)(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.

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS