Difference between revisions of "2017 AMC 10B Problems/Problem 25"
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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}</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>. | ||
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| + | ==See Also== | ||
| + | {{AMC10 box|year=2017|ab=B|num-b=24|num-a=Last Problem}} | ||
| + | {{MAA Notice}} | ||
Revision as of 10:18, 16 February 2017
Contents
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 6 tests be 
. Since the mean of her first 7 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 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 
.
Cheap Solution
By inspection, the sequences 
and 
work, so the answer is .
See Also
| 2017 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Problem 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
