[guahguahmonster]

Oh whoa... I did not realize the person who got the last question was giving the next question. Hah... let me get out my AIME pamphlet and find a good one for you guys...

This was from an AIME test, so the answer is an integer between 0 and 999, inclusive. The answers are probably available online... so no cheating I started with one of the (IMO) easier questions.

There exist unique positive integers x and y that satisfy the equation x^2+84x+2008=y^2. Find x + y.

EDIT: Yeah, you'll want pen and paper for this one. And don't worry, I was able to solve this one with pen and paper (No calculators allowed on AIME) For my curiousity, it'd be great if you posted your work too, so I can compare with how I solved it. And if people get stuck, I'll give you guys a few pointers on how I did it as well...

This post has been edited by guahguahmonster: 25 December 2008 - 04:19 PM

[rdsrds2120]

x^2+84x+2008=y^2. Find x + y.

Wait~wouldn't it equal anypositive number greater than the sqrt(2093)+1? I am asking because you can substitue any number for one of the variables and solve for the other. So I can't really see it being an integral answer, but I think I might be reading the question wrong. I got the 2093 by plgging in 1 for x and solving for y.
Yeah, I am pretty sure I am probably reading the question wrong, hahaha.

This post has been edited by rdsrds2120: 25 December 2008 - 06:11 PM

[GWatt]

guahguahmonster said that there are unique integers for x and y from 0-999 for which that equation works

[rdsrds2120]

oh ok, gotcha

[Gloin]

You can rewrite the equation as (x+42)^2 + 244 = y^2

The final answer must be an even number since x is odd iff (if and only if) y is odd.

You need to find a w^2 + 244 = y^2 (where w = x + 42)
Alot of numbers can be discarded since the least significant digit of any square (y^2) must be in the set [0,1,4,5,6,9] which means that the least significant digit of w^2 must be in the set [0,1,2,5,6,7]. But the least significant digit of w can only be in the intersection of these two sets [0,1,5,6] (since again a square can only have least significant digit equal to a number from the first set) so you need only try the solutions for the following w's.

w=k, w=k+1, w=k+5, w=k+6 (k is any multiple of 10 within the bound)

A solution can be found for w = 60 => x = 18, y = 62 => x+y = 80

This is likely not the best way to do it but it works.

This post has been edited by Gloin: 26 December 2008 - 03:13 AM

[Gloin]

I have a question as well but I was awaiting some discussion about the method of solving the problem

[rdsrds2120]

ok, thanks Gloin, brought back some factoring memories. Good job.

[Gloin]

Calculate (1^1 * 2^2 * 3^3 * ... * 29^29 * 30^30) mod 31
... being the terms in between following the same pattern.

Pen and paper allowed, no calculators though most can't handle numbers this large anyways.
The answer of course being in the range [0..30]

This post has been edited by Gloin: 27 December 2008 - 11:17 AM

[guahguahmonster]

Gloin: Interesting method of solving. Still some brute-forcing involved, but very good elimination tactic based on the units digit of any squared number... I never would have thought of that. How I did it...
(x+42)^2 + 244 = y^2
(x+42)^2 = y^2 - 244
x + 42 = sqrt(y^2 - 244)
Up to here, same as how you've done it. My thought process the diverged:

y^2 obviously must be a perfect square, but because we know x is an integer, y^2-244 must also be a perfect square.
As you had it, we'll let w^2 + 244 = y^2
The difference between any two consecutive squares n^2 and (n+1)^2 is 2n+1, which is odd, so y - w must be even, otherwise the differences could not add up to 244
Factoring 244 (2*2*61) reveals that y - w actually could only be 2 (244 = 121+123).
2w+1=121. so w = 60, y = 62. (And squaring the numbers confirms that 62^2-244=60^2)
You already found that w - 42 = x, so x = 18, arriving at the same answer.

Now I'll think about yours for a bit...

[Gloin]

My first idea was actually to try and factor the number 244 into primes and work with that but I never really went through with the idea or rather, didn't quite get anywhere. Always good to learn new ways though.

[Gloin]

I'm starting to think that maybe my question was a bit too hard. It's really not a difficult problem but it does take alot of time to solve since you have to make so many calculations even if you manage to simplify it using some clever methods.

A hint: A useful fact that I realized when solving this on paper (that I hadn't thought of before I proposed the question) was that using mod 31 simplified the whole solving alot since 31 is one less than a power of 2.

This post has been edited by Gloin: 28 December 2008 - 03:14 AM

[Gloin]

I'm a little disappointed now. I would have expected that someone at least had guessed or cheated their way to an answer.

[rdsrds2120]

eh~I just don't feel like doing 30^30*29^29 on paper just for a forum game, and I dont want to cheat.

[Gloin]

The whole thing is that you can reduce it to a much simpler expression...
For example, 30^30 mod 31 is the same as (-1)^30 mod 31 and since (-1)^30 = 1 you can just discard that factor as it's also 1 mod 31.

[rdsrds2120]

but still, 29*29*29*29*29*29*29*29*29*29*29*29*29*29*29*29*29*29*29*29*29*29*29*29*29*29*29*29*29*28*28*28*28*28*28*28*28*28*28*28*28*28*28*28*28*28*28*28*28*28*28*28*28*28*28*28*28....contd
Is still a lot of multiplication, and a really big number. I definitely would if you would have had them cubed or to the 4th or something like that, but I think that is a little ridiculous.

This post has been edited by rdsrds2120: 30 December 2008 - 04:40 PM