Difference between revisions of "2012 USAJMO Problems/Problem 5"
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| − | The key insight in this | + | 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(mod 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(mod 4)</math> idea. What if we just took <math>2</math> and plugged it in with <math>1006</math>? |
| − | We get 502. | + | 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 01:15, 9 July 2012
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, formal argument
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.
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 | ||
