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

Revision as of 13:36, 28 January 2018

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 $5$ 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 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 negate $95$ 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 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 3 (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.

Solution 4

Since all of the scores are from $91 - 100$ , we can 'subtract' 90 off from all of the scores. Basically, we're looking at the 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 $2 \mod 7$ and $0 \mod 6$ . Trying out a few cases, the only solution possible is 30 (this is from adding numbers 1-10). The sixth test score must be $0 \mod 5$ because $30\equiv 0\mod5$ . The only possible test scores are $95$ and $100$ , so the answer is $\boxed{100}$ .

@@ Line 17: / Line 17: @@
 ==Solution 4==
-Since all of the scores are from <math>91 - 100</math>, we can 'subtract' 90 off from all of the scores. Basically, we're looking at the 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 <math>2 \mod 7</math> and <math>0 \mod 6</math>. Trying out a few cases, the only solution possible is 30 (this is from adding numbers 1-10). The sixth test score must be <math>0 \mod 5</math> because <math>30 = 0\mod5</math>. The only possible test scores are <math>95</math> and <math>100</math>, so the answer is <math>\boxed{100}</math>.
+Since all of the scores are from <math>91 - 100</math>, we can 'subtract' 90 off from all of the scores. Basically, we're looking at the 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 <math>2 \mod 7</math> and <math>0 \mod 6</math>. Trying out a few cases, the only solution possible is 30 (this is from adding numbers 1-10). The sixth test score must be <math>0 \mod 5</math> because <math>30\equiv 0\mod5</math>. The only possible test scores are <math>95</math> and <math>100</math>, so the answer is <math>\boxed{100}</math>.
 ==See Also==
 {{AMC10 box|year=2017|ab=B|num-b=24|after=Last Problem}}
 {{MAA Notice}}

2017 AMC 10B (Problems • Answer Key • Resources)
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