Well, that solution on that forum says:
A) X-Y divided by 5 gives remainder 1, then X-Y=5a+1
B) X+Y divided by 5 gives remainder 2, then X+Y=5b+2
Sum up A) and B): X-Y+X+Y=5a+1+5b+2 -> 2X=5(a+b)+3: when 2X is divided by 5, the remainder is 3. Then, when X is divided by 5, the remainder is 4, as the only values that satisfy 2X=5(a+b)+3 are 4; 9; 14... etc.
Subtract A) from B): X+Y-X+Y=5b+2-5a-1 -> 2Y=5(b-a)+1: when 2X is divided by 5, the remainder is 1. Then, when Y is divided by 5, the remainder is 3, as the only values that satisfy 2Y=5(b-a)+1 are 3; 8; 13... etc.
Now, knowing that:
- X divided by 5 gives remainder 4, and
- Y divided by 5 gives remainder 3
we can calculate the remainder when \(X^4+Y^4\) is divided by 5, and the answer is C.
But I think, that the answer is E, because:
Sum up A) and B): X-Y+X+Y=5a+1+5b+2 -> 2X=5(a+b)+3, when 2X is divided by 5, the remainder is 3. Then, when X is divided by 5, the remainder is 4
or 1.5 (X is not integer), as the values that satisfy 2X=5(a+b)+3 are 4;
6.5; 9;
11.5; 14... etc.
Subtract A) from B): X+Y-X+Y=5b+2-5a-1 -> 2Y=5(b-a)+1, when 2X is divided by 5, the remainder is 1. Then, when Y is divided by 5, the remainder is 3
or 0.5 (Y is not integer), as the values that satisfy 2Y=5(b-a)+1 are 3;
5.5; 8;
10.5; 13... etc.
Then, when \(X^4+Y^4\) will result in decimal without "remainders" when divided by 5.
So, the answer is E unless we are given that X and Y are integers, then the answer will be C.