Difference between revisions of "2012 USAJMO Problems/Problem 5"
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--[[User:Va2010|Va2010]] 11:12, 28 April 2012 (EDT)va2010 | --[[User:Va2010|Va2010]] 11:12, 28 April 2012 (EDT)va2010 | ||
− | == Alternate | + | == Alternate Solution== |
− | Say that the problem is a race track with 2012 spots. To intersect the most, we should get next to each other a lot so the negation is high. As 2012=2^2*503, we intersect at a lot of multiples of 503. | + | Say that the problem is a race track with <math>2012</math> spots. To intersect the most, we should get next to each other a lot so the negation is high. As <math>2012=2^2*503</math>, we intersect at a lot of multiples of <math>503</math>. |
==See also== | ==See also== |
Revision as of 15:29, 2 January 2013
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 |