Difference between revisions of "2012 USAJMO Problems/Problem 5"

(Solution)
m (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. But look at the 2(mod 4) idea. What if we just took 2 and plugged it in. with 1006.
+
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 00:15, 9 July 2012

Problem

For distinct positive integers $a$, $b < 2012$, define $f(a,b)$ to be the number of integers $k$ with $1 \le k < 2012$ such that the remainder when $ak$ divided by 2012 is greater than that of $bk$ divided by 2012. Let $S$ be the minimum value of $f(a,b)$, where $a$ and $b$ range over all pairs of distinct positive integers less than 2012. Determine $S$.

Solution

The key insight in this problem is noticing that when $ak$ is higher than $bk$, $a(2012-k)$ is lower than $b(2012-k)$, except at $2(mod 4)$ residues*. Also, they must be equal many times. $2012=2^2*503$. 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 (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6
All USAJMO Problems and Solutions
Invalid username
Login to AoPS