# 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}$. ==Solution 2 (Cheap Solution)== By inspection, the sequences$ (Error compiling LaTeX. ! Missing $inserted.)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.

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