2006 AMC 12A Problems/Problem 9

Revision as of 21:04, 30 January 2007 by Golddog (talk | contribs) (Solution)

Problem

Oscar buys $13$ pencils and $3$ erasers for $1.00$. A pencil costs more than an eraser, and both items cost a whole number of cents. What is the total cost, in cents, of one pencil and one eraser?

$\mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 12\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 18$

$\mathrm{(E) \ }  20$

Solution

Let the price of a pencil be p and an eraser e.Then 13p + 3e = 100 with p > e > 0. So e >= 1 and p >= 2. Considering the equation 13p + 3e = 100 modulo 3 (remainders when divided by 3) we have p + (0e) = 1 or p = 1. Since p >= 2 possible values for p are {4, 7, 11 ....} Since 13 pencils cost less than 100 cents 13p < 100. 13 x 11 = 101 which is too high so p = 4 or 7.

If p = 4 then 13p = 52 and so 3e = 48 giving e = 16. This contradicts the pencil being more expensive.

The only remaining value for p is 7 and the 13 pencils cost 7 x 13= 91 and so the 3 erasers cost 9 and each eraser cost 9/3 = 3.

One pencil plus one eraser cost 7 + 3 = 10 cents.

Answer A (10)

See also