Difference between revisions of "2017 AIME I Problems/Problem 2"
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The answer is <math>m+n+r+s = 17+31+5+9 = \boxed{62}</math> | The answer is <math>m+n+r+s = 17+31+5+9 = \boxed{62}</math> | ||
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==See Also== | ==See Also== | ||
{{AIME box|year=2017|n=I|num-b=1|num-a=3}} | {{AIME box|year=2017|n=I|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:31, 25 July 2017
Problem 2
When each of ,
, and
is divided by the positive integer
, the remainder is always the positive integer
. When each of
,
, and
is divided by the positive integer
, the remainder is always the positive integer
. Find
.
Solution
Let's tackle the first part of the problem first. We can safely assume:
Now, if we subtract two values:
which also equals
Similarly,
Since
is the only common factor, we can assume that
, and through simple division, that
.
Using the same method on the second half:
Then.
The common factor is
, so
and through division,
.
The answer is
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.