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

Revision as of 08:27, 28 December 2021

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 $S \equiv3 \text{ (mod 7)}$ . 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 $5$ test scores 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 2

First, we find the largest sum of scores which is $100+99+98+97+96+95+94$ which equals $7(97)$ . 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 $7(97), 7(96), 7(95)$ and $7(94)$ which are $679, 672, 665,$ and $658$ respectively. Now in order to get the sum of the first 6 tests, we subtract $95$ from each sum producing $584, 577, 570,$ 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 we have to find which number will give us a number divisible by $5$ when subtracted from $570.$ Since $95$ is the $7th$ test score and all test scores are distinct that only leaves $\boxed{\textbf{(E) } 100}$ .

Solution 3

Since all of the scores are from $91 - 100$ , we can subtract 90 from all of the scores. Basically, we're looking at their units digits (except for 100 - we're looking at 10 in this case). Since the last score was a 95, the sum of the scores from the first six tests must be $3 \pmod 7$ and $0 \pmod 6$ . Trying out a few cases, the only solution possible is 30 (this is from adding numbers 1-10). The sixth test score must be $0 \pmod 5$ because $30\equiv 0\pmod5$ . The only possible test scores are $95$ and $100$ , and $95$ is already used, so the answer is $\boxed{\textbf{(E)}100}$ .

Solution 4 (Working Backwards)

I am going to work backwards to solve this problem. In the $n^{th}$ test, $\sum_{i=1}^{n} i$ (mod n) = 0. In the following table, the tests already taken are in bold, the latest test is underlined. We will work from the row of (mod 7), (mod 6), and (mod 5) to determine the test order by trial and error.

$\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hline & 91 & 92 & 93 & 94 & 95 & 96 & 97 & 98 & 99 & 100 \\ \hline mod 7 & \textbf{0} & \textbf{1} & 2 & \textbf{3} & \underline{\textbf{4}} & \textbf{5} & \textbf{6} & 0 & 1 & \textbf{2} \\ \hline mod 6 & \textbf{1} & \textbf{2} & 3 & \textbf{4} & 5 & \textbf{0} & \textbf{1} & 2 & 3 & \underline{\textbf{4}} \\ \hline mod 5 & \textbf{1} & \textbf{2} & 3 & \underline{\textbf{4}} & 0 & \textbf{1} & \textbf{2} & 3 & 4 & 0 \\ \hline mod 4 & \underline{\textbf{3}} & \textbf{0} & 1 & 2 & 3 & \textbf{0} & \textbf{1} & 2 & 3 & 0 \\ \hline mod 3 & 1 & \textbf{2} & 0 & 1 & 2 & \textbf{0} & \underline{\textbf{1}} & 2 & 0 & 1 \\ \hline mod 2 & 1 & \underline{\textbf{0}} & 1 & 0 & 1 & \textbf{0} & 1 & 0 & 1 & 0 \\ \hline \end{tabular}$

In the $4^{th}$ test, the test score could be 91 or 97.

As you can see, the test score of the $6^{th}$ test is $\boxed{\textbf{(E)}100}$

~isabelchen

Video Solution

https://youtu.be/YFz4bctJYVE - Happytwin

@@ Line 40: / Line 40: @@
 As you can see, the test score of the <math>6^{th}</math> test is <math>\boxed{\textbf{(E)}100}</math>
-~isabelchen
+~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 ==Video Solution==

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