Difference between revisions of "2017 AIME I Problems/Problem 2"
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Doing the same thing for the next three numbers, we get <math>17 + 5 + 31 + 9 = \boxed{62}</math> | Doing the same thing for the next three numbers, we get <math>17 + 5 + 31 + 9 = \boxed{62}</math> | ||
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==Video Solution== | ==Video Solution== |
Revision as of 16:22, 9 March 2021
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 are and , respectively. Since the only common divisor (besides , of course) is , . Dividing all numbers by yields a remainder of for each, so . Thus, .
Solution 2
We know that and where are integers.
Subtracting the first two, the first and third, and the last two, we get and
We know that and must be integers, so all the numbers are divisible by
Factorizing the numbers, we get and We see that all these have a factor of 17, so
Finding the remainder when is divided by we get
Doing the same thing for the next three numbers, we get
~solasky
Video Solution
https://youtu.be/BiiKzctXDJg ~Shreyas S
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.