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

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==Solution 1==
 
==Solution 1==
 
Let the sum of the scores of Isabella's first <math>6</math> tests be <math>S</math>. Since the mean of her first <math>7</math> 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 <math>5</math> integers must be an integer, so the sum of the first <math>5</math> test scores must be an multiple of <math>5</math>, which implies that the <math>6</math>th test score is <math>\boxed{\textbf{(E) } 100}</math>.
 
Let the sum of the scores of Isabella's first <math>6</math> tests be <math>S</math>. Since the mean of her first <math>7</math> 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 <math>5</math> integers must be an integer, so the sum of the first <math>5</math> test scores must be an multiple of <math>5</math>, which implies that the <math>6</math>th test score is <math>\boxed{\textbf{(E) } 100}</math>.
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==Solution 1.1==
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First, we find the largest sum of scores which is <math>100+99+98+97+96+95+94</math> which equals <math>7(98)</math>. Then we find the smallest sum of scores which is <math>91+92+93+94+95+96+97</math> which is <math>7(94). So the possible sums for the 7 test scores so that they provide an integer average are </math>7(98), 7(97), 7(96), 7(95)<math> and </math>7(94)<math> which are </math>686, 679, 672, 665,<math> and </math>658<math> respectively.  Now in order to get the sum of the first 6 tests, we negate </math>95<math> from each sum producing </math>591, 584, 577, 570,<math> and </math>563<math>. Notice only </math>570<math> is divisible by </math>6<math> so, therefore, the sum of the first </math>6<math> tests is </math>570<math>. We need to find her score on the </math>6th<math> test so what number minus </math>570<math> will give us a number divisible by </math>5<math>. Since </math>95<math> is the </math>7th<math> test score and all test scores are distinct that only leaves </math>\boxed{\textbf{(E) } 100}<math>.
  
 
==Solution 2 (Cheap Solution)==
 
==Solution 2 (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>.
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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}$.
 
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.
 
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.
  

Revision as of 09:47, 11 October 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 $6$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.

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
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Problem 24
Followed by
Last Problem
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All AMC 10 Problems and Solutions

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