Difference between revisions of "2017 AMC 10B Problems/Problem 25"
m (→Solution 2 (Cheap Solution)) |
AllenBros713 (talk | contribs) m (→Solution 2 (Cheap Solution)) |
||
Line 10: | Line 10: | ||
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>. | ||
− | 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. | + | 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== | ==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 22:39, 14 March 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 (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.