# Difference between revisions of "2008 AMC 8 Problems/Problem 15"

## Problem

In Theresa's first $8$ basketball games, she scored $7, 4, 3, 6, 8, 3, 1$ and $5$ points. In her ninth game, she scored fewer than $10$ points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than $10$ points and her points-per-game average for the $10$ games was also an integer. What is the product of the number of points she scored in the ninth and tenth games? $\textbf{(A)}\ 35\qquad \textbf{(B)}\ 40\qquad \textbf{(C)}\ 48\qquad \textbf{(D)}\ 56\qquad \textbf{(E)}\ 72$

## Solution

The total number of points from the first $8$ games is $7+4+3+6+8+3+1+5=37$. We have to make this a multiple of $9$ by scoring less than $10$ points. The closest multiple of $9$ is $45$. $45-37=8$ now we have to add a number to get a multiple of 10. The next multiple is $50$ we added $5$, multiplying these together you get $8\cdot5$ is $40$. The answer is $\boxed{B, 40}$

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