Difference between revisions of "2017 AMC 10B Problems/Problem 25"
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==Solution 3== | ==Solution 3== | ||
− | Since all of the scores are from <math>91 - 100</math>, we can 'subtract' 90 off from all of the scores. Basically, we're looking at the units digits except for 100; we're looking at 10 in this case. Since the last score was a 95, the sum of the scores from the first six tests must be <math>3 \mod 7</math> and <math>0 \mod 6</math>. Trying out a few cases, the only solution possible is 30 (this is from adding numbers 1-10). The sixth test score must be <math>0 \mod 5</math> because <math>30\equiv 0\mod5</math>. The only possible test scores are <math>95</math> and <math>100</math>, so the answer is <math>\boxed{\textbf{(E)}100}</math>. | + | Since all of the scores are from <math>91 - 100</math>, we can 'subtract' 90 off from all of the scores. Basically, we're looking at the units digits except for 100; we're looking at 10 in this case. Since the last score was a 95, the sum of the scores from the first six tests must be <math>3 \mod 7</math> and <math>0 \mod 6</math>. Trying out a few cases, the only solution possible is 30 (this is from adding numbers 1-10). The sixth test score must be <math>0 \mod 5</math> because <math>30\equiv 0\mod5</math>. The only possible test scores are <math>95</math> and <math>100</math>, and <math>95</math> is already used, so the answer is <math>\boxed{\textbf{(E)}100}</math>. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=24|after=Last Problem}} | {{AMC10 box|year=2017|ab=B|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:51, 17 January 2020
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 tests be
. Since the mean of her first
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
integers 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
.
Solution 2
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
.
Solution 3
Since all of the scores are from , we can 'subtract' 90 off from all of the scores. Basically, we're looking at the units digits except for 100; we're looking at 10 in this case. 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
.
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by 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 |
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