2017 AMC 10B Problems/Problem 25
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?
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 the CRT, . We also know that , so by inspection, . However, we also have that the mean of the first test scores 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 .
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 .
Since all of the scores are from , we can subtract 90 from all of the scores. 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 .
Solution 4 (Working Backwards)
We work backwards to solve this problem. In the test, (mod n) = 0. In the following table, the tests already taken are in bold, the latest test is underlined. We work from the row of (mod 7), (mod 6), and (mod 5) to determine the test order by trial and error.
In the test, the test score could be 91 or 97.
As you can see, the test score of the test is
Let's denote as , where is her th test score. We start by saying . This results from the fact that her test averages are integers and her test scores are in the range of to . So . Next, we also see that which becomes . The only point of intersection of and is so . Now we can also see that , so , and since , . has to be in the range of and , so the only values in the set that work are and which result in respectively. But since her test scores are all distinct, and is already used for , our answer is .
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