Difference between revisions of "2017 AMC 12B Problems/Problem 21"
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==Solution 2== | ==Solution 2== | ||
Let <math>n</math> be Isabella's average after <math>6</math> tests. <math>6n+95 \equiv 0 \pmod{7}</math>, so <math>n \equiv 4 \pmod{7}</math>. The only integer between <math>90</math> and <math>100</math> that satisfies this condition is <math>95</math>. Let <math>m</math> be Isabella's average after <math>5</math> tests, and let <math>a</math> be her sixth test score. <math>5m+a= 6\cdot95 \equiv 0 \pmod{5}</math>, so <math>a</math> is a multiple of <math>5</math>. Since <math>100</math> is the only choice that is a multiple of <math>5</math>, the answer is <math>\boxed{\textbf{(E) }100}</math>. | Let <math>n</math> be Isabella's average after <math>6</math> tests. <math>6n+95 \equiv 0 \pmod{7}</math>, so <math>n \equiv 4 \pmod{7}</math>. The only integer between <math>90</math> and <math>100</math> that satisfies this condition is <math>95</math>. Let <math>m</math> be Isabella's average after <math>5</math> tests, and let <math>a</math> be her sixth test score. <math>5m+a= 6\cdot95 \equiv 0 \pmod{5}</math>, so <math>a</math> is a multiple of <math>5</math>. Since <math>100</math> is the only choice that is a multiple of <math>5</math>, the answer is <math>\boxed{\textbf{(E) }100}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC12 box|year=2017|ab=B|num-b=20|num-a=22}} | ||
| + | {{MAA Notice}} | ||
Revision as of 21:31, 17 February 2017
Contents
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 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 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
