Difference between revisions of "2012 USAJMO Problems/Problem 5"
(→Solution) |
m (fixing bad LaTeX) |
||
Line 4: | Line 4: | ||
== Solution == | == Solution == | ||
− | The key insight in this problem is noticing that when <math>ak</math> is higher than <math>bk</math>, <math>a(2012-k)</math> is lower than <math> b(2012-k)</math>, except at <math>2 \pmod{4}</math> residues*. Also, they must be equal many times. <math>2012=2^2*503</math>. We should have multiples of <math>503</math>. After trying all three pairs and getting <math>503</math> as our answer, we win. But look at the <math>2\pmod{4} | + | The key insight in this problem is noticing that when <math>ak</math> is higher than <math>bk</math>, <math>a(2012-k)</math> is lower than <math> b(2012-k)</math>, except at <math>2 \pmod{4}</math> residues*. Also, they must be equal many times. <math>2012=2^2*503</math>. We should have multiples of <math>503</math>. After trying all three pairs and getting <math>503</math> as our answer, we win. But look at the <math>2\pmod{4}</math> idea. What if we just took <math>2</math> and plugged it in with <math>1006</math>? |
We get <math>502</math>. | We get <math>502</math>. | ||
--[[User:Va2010|Va2010]] 11:12, 28 April 2012 (EDT)va2010 | --[[User:Va2010|Va2010]] 11:12, 28 April 2012 (EDT)va2010 |
Revision as of 12:34, 18 January 2014
Contents
[hide]Problem
For distinct positive integers ,
, define
to be the number of integers
with
such that the remainder when
divided by 2012 is greater than that of
divided by 2012. Let
be the minimum value of
, where
and
range over all pairs of distinct positive integers less than 2012. Determine
.
Solution
The key insight in this problem is noticing that when is higher than
,
is lower than
, except at
residues*. Also, they must be equal many times.
. We should have multiples of
. After trying all three pairs and getting
as our answer, we win. But look at the
idea. What if we just took
and plugged it in with
?
We get
.
--Va2010 11:12, 28 April 2012 (EDT)va2010
Alternate Solution
Say that the problem is a race track with spots. To intersect the most, we should get next to each other a lot so the negation is high. As
, we intersect at a lot of multiples of
.
See also
2012 USAJMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.