2017 AMC 12B Problems/Problem 21
Contents
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?
Solution 1(FAST)
First, remove all the 90s, since they make no impact. So, we have numbers from to . Then, is the 7th number. Let the sum of the first numbers be . Then, and . We easily solve this as . Clearly, the sum of the first numbers must be . In order for the first numbers to sum to a multiple of , the sixth number must also be a multiple of , since is a multiple of . Thus, the only option is the sixth number is , which gives
~ firebolt360
Solution 2
Let us simplify the problem. Since all of Isabella's test scores can be expressed as the sum of and an integer between and , we rewrite the problem into receiving scores between and . Later, we can add to her score to obtain the real answer.
From this point of view, the problem states that Isabella's score on the seventh test was . We note that Isabella received integer scores out of to . Since is already given as the seventh test score, the possible scores for Isabella on the other six tests are .
The average score for the seven tests must be an integer. In other words, six distinct integers must be picked from set above, and their sum with must be a multiple of . The interval containing the possible sums of the six numbers in S are from to . We must now find multiples of from the interval to . There are four possibilities: , , , . However, we also note that the sum of the six numbers (besides ) must be a multiple of as well. Thus, is the only valid choice.(The six numbers sum to .)
Thus the sum of the six numbers equals to . We apply the logic above in a similar way for the sum of the scores from the first test to the fifth test. The sum must be a multiple of . The possible interval is from to . Since the sum of the five scores must be less than , the only possibilities are and . However, we notice that does not work because the seventh score turns out to be from the calculation. Therefore, the sum of Isabella's scores from test to is . Therefore, her score on the sixth test is . Our final answer is .
Solution 3
Let be Isabella's average after tests. , so . The only integer between and that satisfies this condition is . Let be Isabella's average after tests, and let be her sixth test score. , so is a multiple of . Since is the only choice that is a multiple of , the answer is .
Solution 4
Let be the total sum of Isabella's first five test scores, and let be her score on the sixth test. It follows that , , and , since at each step, her average score was an integer. Using the last equivalence, , so we have a system of equivalences for . Solving this using the Chinese Remainder Theorem, we get .
Now let's put a bound on . Using the given information that each test score was a distinct integer from to inclusive and that the seventh score was 95, we get . Since , we get . Therefore,
The last preparation step will involve calculating all the possible test scores . Here they are: . This means that . Note that is not in the previous list because it corresponds to a score of , which we cannot have.
We must have , and using the possible values we found for and , the only two that sum to are and . This corresponds to an value of , so the answer is .
Solution 5
We know that Isabella's score on the 7th test is 95, and we work backwards. Let denote the average of the first scores. We are given that all are integers. If the is different from the , then the last score must be a multiple of 7 away from . However, and clearly cannot be since they are outside the range of 91 to 100. Thus, must be 95. Additionally, cannot be 95 since that would mean the sixth score is 95, which isn't possible since scores can't repeat (also, it's not an answer choice). Intuitively, should be close to 95 due to the constraint on the range of scores. We can try 94 and 96 for . Again, the sixth score must be a multiple of 6 (preferably only one multiple) away from . The only one that works is , and we can check that the rest don't work since they exceed the range. The answer is .
- Lingjun :)
Solution 6
Say the sum of the first 6 scores is . Then, we know that . Thus, . Additionally, since the average of these 6 scores was an integer, we know that . We find that the smallest to satisfy both of these inequalities is 24 (series below). To get to the next one, we have to add so as not to ruin the moduli.
Next, we know that is between and . It can't take on either value as all the scores have to be different. Now, we calculate the first that satisfies both moduli equations:
works while does not (as it hits the upper limit of 600), and we find that . So, the sum of the first 6 scores is 570. We also know that the sum of the first 5 scores is a multiple of 5. Since , the 6th score must be a multiple of 5. 95 is taken, so the only possibility .
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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 12 Problems and Solutions |
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