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. Unknown error_msg)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.

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
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 10 Problems and Solutions

Page

Toolbox

Search

2017 AMC 10B Problems/Problem 25

Contents

Problem

Solution 1

Solution 1.1

See Also