<p>@funsummer: Lol! Actually, the timings in my part of the world are like GMT + 5, so it’s not hardcore. 
Also, how do you prove max (a - b) = 4 using AM-GM?
Guys, I’m loving this thread so far. :)</p>
<p>Rewrite (a^2 + b^2)/(a-b) as (a-b) + 2ab/(a-b) = (a-b) + 16/(a-b) = S.</p>
<p>Apply AM-GM and you get S/2 >= sqrt(16) → S >= 8. Equality occurs if and only if a-b = 4, and you can show that there is a solution that satisfies.</p>
<p>Let a - b = x, 16/(a - b) = y</p>
<p>By AM-GM, we have (x+y)/2 >= sqrt(xy).
Just plug that in and you get a - b + 16/(a - b) >= 8.</p>
<p>omg guys stop sniping me :(</p>
<p>I guess we missed attempting this one:
“Nine different, non-zero integers are placed into a 3x3 grid such that, for every row and column, the sum of the third number and the product of the first two numbers (reading left to right, top to bottom) is a constant. What is the minimum possible sum of the absolute values of the nine integers?”
Is there a quick solution to this?</p>
<p>@Kyrix1: I didn’t quite follow from here: " Using a well-known fact and the fact that a has less than 2011 divisors, we know that A <=2010. If A=2010, the inequality and the fact that B is an integer forces B >=2010 and thus A <= B. If A < 2010, we have the same result (the proof is pretty intuitive; basically multiplying a number less than 2010, say X, by 2011/2012 can’t get you less than or equal to X-1). A <= B implies, as we well know, a | b." 
How did you figure out that A<= 2010 and the subsequent stuff?</p>
<p>The “well-known fact” is here: [url=<a href=“http://mathschallenge.net/library/number/number_of_divisors]mathschallenge.net[/url”>mathschallenge.net]mathschallenge.net[/url</a>]
It easily follows then that since A is an exponent A <= 2010.</p>
<p>Now let A = 2010. Plug that into the inequality, and you see that 2009.000(something not 0) <= B. That means B >= 2010 because B must be an integer.</p>
<p>Now let A < 2010. You can prove from there that the the ceiling function of the left hand side is always greater than or equal to A; in other words, A <= B.</p>
<p>@yellowcat429, there is a somewhat quick solution that involves only a little guess-and-check. I made the problem up while writing practice AIME’s…hopefully there’s no mistake or ambiguity in the solution.</p>
<p>@rspence: What solution do you get?</p>
<p>1 4 -4
3 2 -6
-3 -8 24</p>
<p>Minimal sum is 1+4+4+3+2+6+3+8+24 = 55.</p>
<p>You can solve it by proving that this “constant” must be zero. Then just optimize the upper left 2x2 square.</p>
<p>Nice! 
I’m sorry for dragging this, but how do you know that you would get the minimum sum if the constant is zero? (I can see that once we know this, figuring out the individual terms won’t be much of a problem.)</p>
<p>Because we can show that the constant has to be zero. Suppose the numbers are</p>
<p>a b -ab+k
c d -cd+k
-ac+k -bd+k abcd+k</p>
<p>Then (-ab+k)(-cd+k) = (-ac+k)(-bd+k). Expanding we get</p>
<p>abcd - k(ab + cd) + k^2 = abcd - k(ac + bd) + k^2</p>
<p>k(ab + cd) = k(ac + bd)</p>
<p>k(ab + cd - ac + bd) = 0 → k(a-d)(b-c) = 0.</p>
<p>Since a,b,c,d are distinct, a-d and b-c cannot equal zero, hence k = 0.</p>
<p>Once you know this constant is zero, brute force a bit…</p>
<p>^Thanks! :)</p>
<p>Here’s another question that doesn’t require much hardcore math. </p>
<p>A group of nine friends go to play a game of Ten Pin Bowling. They each need to hire a pair of shoes specially designed for the surface they play on. Unfortunately, the bowling alley does not have the correct size shoe for all the friends. The shoe sizes available are 3 pairs of size 6, 2 pairs of size 8, 2 pairs of size 9 and 3 pairs of size 11. The shoe sizes of the nine friends are 5, 6, 6, 6, 7, 7, 9, 10, 10.
What is the maximum number of friends who will receive at most one size too big?</p>
<p>Arrange in order:</p>
<p>6 6 6 8 8 9 9 11 11 11 ← bowling alley’s shoes
5 6 6 6 7 7 9 10 10</p>
<p>The two friends with 10’s get 11’s, 9 gets 9, the two 7’s get 8’s, the three 6’s get 6. The 5 gets a size at least one too big.</p>
<p>I think that’s the optimal solution, proving it looks ugly.</p>
<p>Yeah, I too reached the conclusion that the size 5 person would receive a 6, the two 7’s would get the two 8’s and the two 10’s would get the two 11’s. That makes the maximum number of friends receiving at most one size too big = 1+2+2 = 5. However, the correct ans for the maximum number is 8! That left me confounded. </p>
<p>(The question doesn’t really ask for a proof. We just have to determine the maximum number.)</p>
<p>Look at the way Rspence lined up the info. Then move the kid with size 5 to the end of the line, gving him a size 11 shoe. Everyone else moves over one spot to the left, receiving either their proper size or one size bigger.</p>
<p>^That still doesn’t give 8 friends with at most one size too big.</p>
<p>Oh, you are right! I read it as “within one size”…my mistake – but if the answer key says 8, maybe it was their mistake too.</p>
<p>BTW, I think that not seeing this solution is a sign of good ethics: you probably didn’t think it was fair to stick the one kid with such a lousy fit.</p>
<p>Yeah, I’m not sure what “at most one size too big” means. </p>
<p>Although I just realized that if you move the bottom column one to the right, you obtain this:</p>
<p>6 6 6 8 8 9 9 11 11 11
–5 6 6 6 7 7 9 10 10</p>
<p>In this case, eight people have a shoe that is too big, but only seven people have a shoe that is exactly one size too big.</p>
<p>Here’s an interesting problem:</p>
<p>The numbers 1,2,…,9 are written on a blackboard. Two players play a game in which the first player takes a number and erases it, the second player takes a number and erases it, then the first player, etc. The goal is to have exactly three numbers that add up to 15. Find a winning strategy for one of the players.</p>