Difference between revisions of "2017 AIME I Problems/Problem 2"
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The common factor is <math>31</math>, so <math>n=31</math> and through division, <math>s=9</math>. | The common factor is <math>31</math>, so <math>n=31</math> and through division, <math>s=9</math>. | ||
− | The answer is <math>m+n+r+s = 17+31+5+9 = \boxed{ | + | The answer is <math>m+n+r+s = 17+31+5+9 = \boxed{62}</math> |
~IYN~ | ~IYN~ |
Revision as of 20:42, 5 June 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
~IYN~
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.