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

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<math>\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100</math>
 
<math>\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100</math>
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==Solution 1==
 +
Let the sum of the scores of Isabella's first 6 tests be <math>S</math>. Since the mean of her first 7 scores is an integer, then <math>S + 95 \equiv 0 \text{ (mod 7)}</math>, or  <math>S \equiv 3 \text{ (mod 7)}</math>. Also, <math>S \equiv 0 \text{ (mod 6)}</math>, so by CRT, <math>S \equiv 24 \text{ (mod 42)}</math>. We also know that <math>91 \cdot 6 \leq S \leq 100 \cdot 6</math>, so by inspection, <math>S = 570</math>. 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 <math>6th</math> test score is <math>\boxed{\textbf{(E) } 100}</math>.
  
 
==Cheap Solution==
 
==Cheap Solution==
  
 
By inspection, the sequences <math>91,93,92,96,98,100,95</math> and <math>93,91,92,96,98,100,95</math> work, so the answer is <math>\boxed{\textbf{(E) } 100}</math>.
 
By inspection, the sequences <math>91,93,92,96,98,100,95</math> and <math>93,91,92,96,98,100,95</math> work, so the answer is <math>\boxed{\textbf{(E) } 100}</math>.

Revision as of 10:08, 16 February 2017

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 $6th$ test score is $\boxed{\textbf{(E) } 100}$.

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 $\boxed{\textbf{(E) } 100}$.

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