2017 AIME I Problems/Problem 2
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 work on both parts of the problem separately. First, We take the difference of and , and also of and . We find that they are and , respectively. Since the greatest common divisor of the two differences is (and the only one besides one), it's safe to assume that .
Then, we divide by , and it's easy to see that . Dividing and by also yields remainders of , which means our work up to here is correct.
Doing the same thing with , , and , the differences between and and and are and , respectively. Since the only common divisor (besides , of course) is , . Dividing all numbers by yields a remainder of for each, so . Thus, .
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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