# BasilMarket

## Chat Forum Talk about topics not related to MapleStory

Level 31 Renegades Demon Slayer 2
Let x and y be two-digit integers such that y is obtained by reversing the digits of x. The integers x and y satisfy x^2-y^2 = m^2 for some positive integer m.
What is x+y+m?

Answer, and please explain to me how you got it... thank you!

# Replies

I'll give you a hint, x or y are factors of 9, oh and x will be higher than y.
Nov 18 2012
Level 136 Zenith Dark Knight
91 and 18? @DayServant sorry if I ruined the secret
Nov 18 2012
Level 31 Renegades Demon Slayer 2
91 and 18? @DayServant sorry if I ruined the secret[/quote]

it can also be 27 and 72...
Nov 18 2012
Level 136 Zenith Dark Knight
I honestly didn't think of that, my excuse is that it's almost 4am
Nov 18 2012
I'm not sure as to what exactly you are trying to find.

Btw I'll have to redo the calculations.
Nov 18 2012
Level 136 Zenith Dark Knight
Hold up, aren't we suppose to take 18^2 from 81^2? or did I read this wrong?
Nov 18 2012
x=11, y=11

11^2 - 11^2 = 0

0^2 = 0

11+11+0=22

[lol. but seriously you haven't given me enough information.]
Nov 18 2012
Level 200 Mardia Corsair
theres at least 9 unique solutions here, because there is not enough restrictions
where x and y could be the same
11, 22, 33, 44, 55, 66, 77, 88, 99
Nov 18 2012
Level 200 Broa Night Lord bombinator
oh my bad, wasnt thinking of the other digits
Nov 18 2012
after about 30 mins... finished a program that gets the answer by the use of bruteforce (trial and error), nevertheless it outputs the answer. I'm sure there's a way with limits to find the answer..

#include <iostream>
#include <math.h>
using namespace std;
int main(){
int iNum1 =2, iNum2=1;
int iResult1, iResult2;
int iSqrt1, iSqrt2;
double dAns;
int aNum1[2];
int aNum2[2];
for(int c=0;c<=101;c++)
{
//cout <<"iNum2 = "<< iNum2;
//cout <<"tiNum1 = "<< iNum1 << endl;
if(iNum2==10)
{
iNum1++;
iNum2=0;
}
iResult1 = 0;
iResult2 = 0;
int aNum1[2]={iNum1,iNum2};
int aNum2[2]={iNum2,iNum1};
//concatenates the 2 digit number
iResult1 = ((aNum1[0])*10)+aNum1[1];
//cout << "iResult1= " << iResult1 << endl;
iResult2 = ((aNum2[0])*10)+aNum2[1];
// cout << "iResult2= " << iResult2 << endl;

if((iResult1 <= iResult2))
{
cout << "empty setn";
iNum2++;
}
else if(iResult2 <= 10)
{
cout << "empty setn";
iNum2++;
}else
{
iSqrt1 = (iResult1) * (iResult1);
iSqrt2 = (iResult2) * (iResult2);
dAns = iSqrt1 - iSqrt2;
cout << iResult1<< " - " <<iResult2<<" = " << sqrt(dAns) << endl;
iNum2++;
}
}
return 0;
}
[/quote]

btw the answer is 154... (65+56+33)
Nov 18 2012