Difference between revisions of "2017 AMC 10B Problems/Problem 25"
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First, we find the largest sum of scores which is <math>100+99+98+97+96+95+94</math> which equals <math>7(97)</math>. Then we find the smallest sum of scores which is <math>91+92+93+94+95+96+97</math> which is <math>7(94)</math>. So the possible sums for the 7 test scores so that they provide an integer average are <math>7(97), 7(96), 7(95)</math> and <math>7(94)</math> which are <math>679, 672, 665,</math> and <math>658</math> respectively. Now in order to get the sum of the first 6 tests, we negate <math>95</math> from each sum producing <math>584, 577, 570,</math> and <math>563</math>. Notice only <math>570</math> is divisible by <math>6</math> so, therefore, the sum of the first <math>6</math> tests is <math>570</math>. We need to find her score on the <math>6th</math> test so what number minus <math>570</math> will give us a number divisible by <math>5</math>. Since <math>95</math> is the <math>7th</math> test score and all test scores are distinct that only leaves <math>\boxed{\textbf{(E) } 100}</math>. | First, we find the largest sum of scores which is <math>100+99+98+97+96+95+94</math> which equals <math>7(97)</math>. Then we find the smallest sum of scores which is <math>91+92+93+94+95+96+97</math> which is <math>7(94)</math>. So the possible sums for the 7 test scores so that they provide an integer average are <math>7(97), 7(96), 7(95)</math> and <math>7(94)</math> which are <math>679, 672, 665,</math> and <math>658</math> respectively. Now in order to get the sum of the first 6 tests, we negate <math>95</math> from each sum producing <math>584, 577, 570,</math> and <math>563</math>. Notice only <math>570</math> is divisible by <math>6</math> so, therefore, the sum of the first <math>6</math> tests is <math>570</math>. We need to find her score on the <math>6th</math> test so what number minus <math>570</math> will give us a number divisible by <math>5</math>. Since <math>95</math> is the <math>7th</math> test score and all test scores are distinct that only leaves <math>\boxed{\textbf{(E) } 100}</math>. | ||
− | ==Solution | + | ==Solution 3 (Cheap Solution)== |
By inspection, the sequences <math>91,93,92,96,98,100,95</math> and <math>93,91,92,96,98,100,95</math> work, so the answer is <math>\boxed{\textbf{(E) } 100}</math>. | By inspection, the sequences <math>91,93,92,96,98,100,95</math> and <math>93,91,92,96,98,100,95</math> work, so the answer is <math>\boxed{\textbf{(E) } 100}</math>. |
Revision as of 00:58, 6 November 2017
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 negate
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 what number minus
will give us a number divisible by
. Since
is the
test score and all test scores are distinct that only leaves
.
Solution 3 (Cheap Solution)
By inspection, the sequences and
work, so the answer is
.
Note: A method of finding this "cheap" solution is to create a "mod chart", basically list out the residues of 91-100 modulo 1-7 and then finding the two sequences should be made substantially easier.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.