Difference between revisions of "2002 AIME II Problems/Problem 7"
m |
|||
Line 16: | Line 16: | ||
Thus, there are no restrictions on <math>k</math> in <math>\pmod{3}</math>. | Thus, there are no restrictions on <math>k</math> in <math>\pmod{3}</math>. | ||
− | + | It is easy to see that only one of <math>k</math>, <math>k+1</math>, and <math>2k+1</math> is divisible by <math>5</math>. So either <math>k, k+1, 2k+1 \equiv 0 \pmod{25}</math>. | |
Thus, <math>k \equiv 0, 24, 12 \pmod{25}</math>. | Thus, <math>k \equiv 0, 24, 12 \pmod{25}</math>. |
Revision as of 00:34, 27 December 2014
Problem
It is known that, for all positive integers ,

Find the smallest positive integer such that
is a multiple of
.
Solution
is a multiple of
if
is a multiple of
.
So
.
Since is always odd, and only one of
and
is even, either
.
Thus, .
If , then
. If
, then
. If
, then
.
Thus, there are no restrictions on in
.
It is easy to see that only one of ,
, and
is divisible by
. So either
.
Thus, .
From the Chinese Remainder Theorem, . Thus, the smallest positive integer
is
.
See also
2002 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.