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What is the remainder when x^4 + y^4 divided by 5?

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cledgard

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20 Oct 2016, 12:23

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Sunchaser20

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.

The problem does not need to specify that the variables are integers; if the question asks for a remainder, then there must be a remainder. The remainder is always an integer; if the answer is not an integer, then it is not a remainder; therefore it is not a valid answer.

Does this help?

cledgard

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20 Oct 2016, 12:32

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skpMatcha

Sure thanks for the explanation.

what threw me off is.. the remainder is 4 or 1.5(X is not an integer). I was expecting by your explanation , when (a+b is even ). So got confused but now realized that they are one and the same

I hope real GMAT doesnt ask such heavy qns

Quote:

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.

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20 Oct 2016, 14:02

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mdfrahim

What is the remainder when x^4 + y^4 divided by 5?

(1) x - y divided by 5 gives remainder 1
(2) x + y divided by 5 gives remainder 2

Show Spoiler

I got this from some other forum. Here is the link
https://www.manhattangmat.com/forums/x4- ... t7277.html
I was not able to undersand the solution at all. I am sure on this forum I will definitely be able to get an answer which I am looking for.

It is pure algebra ....

2(x^4+y^4) = (x^2+y^2)^2 + (x^2-y^2)^2 and 2(x^2+y^2) = (x+y)^2 + (x-y)^2

each alone is insuff
together

(x-y) = 5m+1, x+y = 5n+2

(x^2 - y^2)^2 = {(x-y)(x+y)]^2 = (5m+1)^2* (5n+2)^2 = (25m+10m+1)*(25n+10n+4) = ...... only 1*4 is not multiple of 5 and giver r= 4

(x+y)^2 + (x-y)^2 = 25m+10m+1+25n+10n+4 = a multiple of 5 thus (x^2+y^2)^2 is multiple of 5

thus 2(x^4+y^4) always gives a remainder of 4 when divided by 5 therefore x^4+y^4 gives a remainder of 2 when divided by 5

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The remainder upon devision by 5 can be found by seeing the units digit of a number. The remainder can be 0.1.2.3.4

I tried the units digit approach.

x-y div by 5 gives 1 as remainder.

find two integers who upon subtraction give 1 or 6 as units digit.

4-3 gives 1
6-5 gives 1 as remainder
9-3 6 as remainder etc

we can raise them to 4th power in each case and check the unites digit of the sum x4+y4

4,3 will give 1 as units digit and 6+1 =7 as units digit of sum giving 2 as remainder
6-5 will give 1 as remainder of subtraction and of the sum 1 as remainder upon div by 5

statement 1 is insufficient

Similarity stat 2

we need to sum two number whose units digit gives us 2 as units digit

9+3 gives 2
7+5 gives 2

when we raise these to 4th power we get 2 as remainder in the first case and 1 in the second case insufficient

now looking at both statements 9 and 3 as units digits satisfies both 9-3 =6 for first statement and 9+3 for second.

other combinations don't satisfy both statements.

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