# Difference between revisions of "2017 AMC 10B Problems/Problem 25"

## Problem

Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100$, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95$. What was her score on the sixth test? $\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$

## Solution 1

Let the sum of the scores of Isabella's first $6$ tests be $S$. Since the mean of her first $7$ scores is an integer, then $S + 95 \equiv 0 \text{ (mod 7)}$, or $S \equiv 3 \text{ (mod 7)}$. Also, $S \equiv 0 \text{ (mod 6)}$, so by CRT, $S \equiv 24 \text{ (mod 42)}$. We also know that $91 \cdot 6 \leq S \leq 100 \cdot 6$, so by inspection, $S = 570$. However, we also have that the mean of the first $5$ integers must be an integer, so the sum of the first $5$ test scores must be an multiple of $5$, which implies that the $6$th test score is $\boxed{\textbf{(E) } 100}$.

## Solution 1.1

First, we find the largest sum of scores which is $100+99+98+97+96+95+94$ which equals $7(98)$. Then we find the smallest sum of scores which is $91+92+93+94+95+96+97$ which is $7(94)$. So the possible sums for the 7 test scores so that they provide an integer average are $7(98), 7(97), 7(96), 7(95)$ and $7(94)$ which are $686, 679, 672, 665,$ and $658$ respectively. Now in order to get the sum of the first 6 tests, we negate $95$ from each sum producing $591, 584, 577, 570,$ and $563$. Notice only $570$ is divisible by $6$ so, therefore, the sum of the first $6$ tests is $570$. We need to find her score on the $6th$ test so what number minus $570$ will give us a number divisible by $5$. Since $95$ is the $7th$ test score and all test scores are distinct that only leaves $\boxed{\textbf{(E) } 100}$.

## Solution 2 (Cheap Solution)

By inspection, the sequences $91,93,92,96,98,100,95$ and $93,91,92,96,98,100,95$ work, so the answer is $\boxed{\textbf{(E) } 100}$. Note: A method of finding this "cheap" solution is to create a "mod chart", basically list out the residues of 91-100 modulo 1-7 and then finding the two sequences should be made substantially easier.

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