2017 AIME I Problems/Problem 2
Contents
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,
.
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 | ||
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