Difference between revisions of "2012 USAJMO Problems/Problem 5"
(→Solution) |
(→Solution) |
||
| Line 4: | Line 4: | ||
== Solution == | == Solution == | ||
| − | The key insight in this problemo is noticing that when ak is higher then bk, a(2012-k) is lower than b(2012-k), except at 2(mod 4) residues*. Also, they must be equal quite a lot. 2012=2^2*503. We should have multiples of 503. After trying all three pairs and getting 503 as our answer, we win. | + | The key insight in this problemo is noticing that when ak is higher then bk, a(2012-k) is lower than b(2012-k), except at 2(mod 4) residues*. Also, they must be equal quite a lot. 2012=2^2*503. We should have multiples of 503. After trying all three pairs and getting 503 as our answer, we win. But look at the 2(mod 4) idea. What if we just took 2 and plugged it in. with 1006. |
| − | + | We get 502. | |
| + | --[[User:Va2010|Va2010]] 11:12, 28 April 2012 (EDT)va2010 | ||
| + | |||
== Alternate, formal argument== | == 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. | 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. | ||
Revision as of 10:43, 28 April 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 problemo is noticing that when ak is higher then bk, a(2012-k) is lower than b(2012-k), except at 2(mod 4) residues*. Also, they must be equal quite a lot. 2012=2^2*503. We should have multiples of 503. After trying all three pairs and getting 503 as our answer, we win. But look at the 2(mod 4) idea. What if we just took 2 and plugged it in. with 1006. We get 502.
--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 | ||
