Difference between revisions of "2017 AMC 12B Problems/Problem 21"
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==Solution 1== | ==Solution 1== | ||
− | Let us simplify the problem. Since all of Isabella's test scores can be expressed as the sum of 90 and an integer between 1 and 10, we rewrite the problem into receiving scores between 1 and 10. Later, we can add 90 to her score to obtain the real answer. | + | Let us simplify the problem. Since all of Isabella's test scores can be expressed as the sum of <math>90</math> and an integer between <math>1</math> and <math>10</math>, we rewrite the problem into receiving scores between <math>1</math> and <math>10</math>. Later, we can add <math>90</math> to her score to obtain the real answer. |
− | From this point of view, the problem states that | + | From this point of view, the problem states that Isabella's score on the seventh test was <math>5</math>. We note that Isabella received <math>7</math> integer scores out of <math>1</math> to <math>10</math>. Since <math>5</math> is already given as the seventh test score, the possible scores for Isabella on the other six tests are <math>S={1,2,3,4,6,7,8,9,10}</math>. |
− | The average score for the seven tests must be an integer. In other words, six distinct integers must be picked from set S above, and their sum with 5 must be a multiple of 7. The interval containing the possible sums of the six numbers in S are from | + | The average score for the seven tests must be an integer. In other words, six distinct integers must be picked from set <math>S</math> above, and their sum with <math>5</math> must be a multiple of <math>7</math>. The interval containing the possible sums of the six numbers in S are from <math>1 +2+3+4+6+7=23</math> to <math>4+6+7+8+9+10=44</math>. We must now find multiples of <math>7</math> from the interval <math>23+5 = 28</math> to <math>44+5=49</math>. There are four possibilities: <math>28</math>, <math>35</math>, <math>42</math>, <math>49</math>. |
− | However, we also note that the sum of the six numbers (besides 5) must be a multiple of 6 as well. Thus, 35 is the only valid choice.(The six numbers sum to 30.) | + | However, we also note that the sum of the six numbers (besides <math>5</math>) must be a multiple of <math>6</math> as well. Thus, <math>35</math> is the only valid choice.(The six numbers sum to <math>30</math>.) |
− | Thus the sum of the six numbers equals to 30. 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 5. The possible interval is from | + | Thus the sum of the six numbers equals to <math>30</math>. 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 <math>5</math>. The possible interval is from <math>1+2+3+4+6=16</math> to <math>6+7+8+9+10=40</math>. Since the sum of the five scores must be less than <math>30</math>, the only possibilities are <math>20</math> and <math>25</math>. However, we notice that <math>25</math> does not work because the sixth score turns out to be <math>5</math> from the calculation. Therefore, the sum of Isabella's scores from test <math>1</math> to <math>5</math> is <math>20</math>. Therefore, her score on the sixth test is <math>10</math>. |
− | Our final answer is 10+90= | + | Our final answer is <math>10+90= \boxed{\textbf{(E) }100}</math>. |
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==Solution 2== | ==Solution 2== | ||
Let <math>n</math> be Isabella's average after 6 tests. <math>6n+95=0 (mod 7)</math>, so <math>n=4 (mod 7)</math>. The only integer between 90 and 100 that satisfies this condition is 95. Let <math>m</math> be Isabella's average after 5 tests, and let <math>a</math> be her sixth test score. <math>5m+a=95</math>, so <math>a</math> is a multiple of 5. Since 100 is the only choice that is a multiple of 5, the answer is 100=<math>D</math>. | Let <math>n</math> be Isabella's average after 6 tests. <math>6n+95=0 (mod 7)</math>, so <math>n=4 (mod 7)</math>. The only integer between 90 and 100 that satisfies this condition is 95. Let <math>m</math> be Isabella's average after 5 tests, and let <math>a</math> be her sixth test score. <math>5m+a=95</math>, so <math>a</math> is a multiple of 5. Since 100 is the only choice that is a multiple of 5, the answer is 100=<math>D</math>. |
Revision as of 20:19, 17 February 2017
Problem 21
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. Was was her score on the sixth test?
Solution 1
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 sixth 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 2
Let be Isabella's average after 6 tests. , so . The only integer between 90 and 100 that satisfies this condition is 95. Let be Isabella's average after 5 tests, and let be her sixth test score. , so is a multiple of 5. Since 100 is the only choice that is a multiple of 5, the answer is 100=.