Difference between revisions of "2017 AMC 10B Problems/Problem 25"
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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>3 \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>, and <math>95</math> is already used, so the answer is <math>\boxed{\textbf{(E)}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>3 \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>, and <math>95</math> is already used, so the answer is <math>\boxed{\textbf{(E)}100}</math>. | ||
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+ | ==Video Solution== | ||
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+ | https://youtu.be/YFz4bctJYVE - Happytwin | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=24|after=Last Problem}} | {{AMC10 box|year=2017|ab=B|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:56, 18 June 2020
Problem
Last year Isabella took math tests and received different scores, each an integer between and , inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was . What was her score on the sixth test?
Solution 1
Let the sum of the scores of Isabella's first tests be . Since the mean of her first scores is an integer, then , or . Also, , so by CRT, . We also know that , so by inspection, . However, we also have that the mean of the first integers must be an integer, so the sum of the first test scores must be an multiple of , which implies that the th test score is .
Solution 2
First, we find the largest sum of scores which is which equals . Then we find the smallest sum of scores which is which is . So the possible sums for the 7 test scores so that they provide an integer average are and which are and respectively. Now in order to get the sum of the first 6 tests, we subtract from each sum producing and . Notice only is divisible by so, therefore, the sum of the first tests is . We need to find her score on the test so we have to find which number will give us a number divisible by when subtracted from Since is the test score and all test scores are distinct that only leaves .
Solution 3
Since all of the scores are from , 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 and . Trying out a few cases, the only solution possible is 30 (this is from adding numbers 1-10). The sixth test score must be because . The only possible test scores are and , and is already used, so the answer is .
Video Solution
https://youtu.be/YFz4bctJYVE - Happytwin
See Also
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 |
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