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

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.

Another approach:
x - y mod 5 = 1
x + y mod 5 = 2
When summing them it becomes
2x mod 5 = 3 = 8 => x mod 5 = 4
When subtracting first from second it becomes
2y mod 5 = 1 = 6 => y mod 5 = 3

As it seen; equation holds.

Thus,
x^4 + y^4 mod 5 = 4^4 + 3^4 mod 5 = 2
So we can solve it by getting them together.

Agree with you, unless it is mentioned that X and Y are integers, ans should be E.

If anyone thinks otherwise, please elaborate.

In my approach; if it is not provided that x and y integer; the answer is E.

maliyeci, I don't understand your approach at all and neither I have ever seen this kind of approach in my entire life (i saw this approach only in GMAT forums only).
Can you please tell me from where I can learn this or pls. explain me a bit in detail, if you feel its practically possible. Will you mind if I send you a PM regarding this.

If there is a subject that I can help you and someone else. I will be very proud to help. In Turkey, we have a very high level math education. In high school we learnt calculus. In college, learnt it again. We learnt trigonometry, combinatorics and other high level subject again and again. So we have very good maths. I do not know how to teach about it.

I guess this explains it quite well.

Thanks maliyeci.Good to know the level of Maths is so high in Turkey.
I am from India and believe it that the level of maths is too high here in India also.I think that there are a lot of people from india in this forum which will agree to me.Infact I consider that maths is the strongest poin for Indians in gmat.
in case of this particular technique , i think that it sure might be taught at the college leve but i am not aware of it since i never got a chance to study this.
Anyways i wll try to dig this question more and ask you for any help if required.

Always as I am here.

Guys, I just heard about remainder theorm today can you please explain me this statement ?

Sum up A) and B):



It is clear till above statement. Then part went right above my head, can you please walk me through this...

Ok, lets just simplify this part. We know, that when 2X is divided by 5, the remainder is 3, i.e. 2X=5n+3 (actually, n=a+b, if we want to link it with the main question, but for simplicity it is better to use only one variable). Now, if we want to know what will be the remainder when X is divided by 5, we can use the equation 2X=5n+3, dividing it by 2. As n is integer, there are 2 possibilities: n is odd or n is even.
1. n is odd, i.e. n=2m+1 (m is integer), then \(X=\frac{2X}{2}=\frac{5n+3}{2}=\frac{5*(2m+1)+3}{2}=\frac{5*2m+5+3}{2}=5m+\frac{8}{2}\), which means, that when X is divided by 5, the remainder is 4, which is the case, considered by
2. n is even, i.e. n=2m (m is integer), then \(X=\frac{2X}{2}=\frac{5n+3}{2}=\frac{5*2m+3}{2}=5m+\frac{3}{2}\), which means, that when X is divided by 5, the "remainder" is 3/2

Hey thanks for explaining, I still have few doubts..



So does that mean , you have not considered (a+b) to be even in this case as stated below? why dint I get a remainder of 1.5 in the above case ? I think I understood but you know I am not there yet

Well, in the first part of my initial post I was explaining in details the solution from the manhattangmat.com forum. And, yes, that solution doesn't consider (a+b) to be even, which I have pointed out in the 2nd part of my post, starting after the words "But I think, that the answer is E, because:"
Please, feel free to ask any questions

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

What is the remainder when X^4 + Y^4 is divided by 5

1. When X-Y is divided by 5 remainder is 1
2. When X+Y is divided by 5 remainder is 2

This question has been treated in another thread, but I simply cannot filter
out the posts for reference.

The approach used in one of the post:
Each statement alone is not SUFFICIENT.
Let X - Y = 5A + 1
X + Y = 5B + 2
2X = 5(A+B) + 3
2Y = 5(B-A) + 1
16 (X4 + Y4) = {5(A+B)+3}4 + {5(B-A)+1}4
= (5P+3)4 + (5Q+1)4
= 81 + 1 + 5R
= 82 + 5R
X4 + Y4 = 5 + (5R+2)/16
Since L.H.S is multiple of 16, R.H.S. should also be multiple of 16(since R.H.S cannot be a fraction.).
Lowest value of R for which R.H.S. is an integer is R=6
X4 + Y4 = 82
Remainder of (X4 + Y4)/5 = 2
Hence C

Could anyone assist to especially with the highlighted part. Thanks

The method you posted seems to me to be a bit too contrived and lengthy, though it may work for some people. On the GMAT, it may be better to test cases and visualize possibilities rather than perform too much of algebra that you would be prone to make errors in under a time crunch. The following method may seem pretty contrived too, but in fact it's quite intuitive - you just need to test obvious cases. This approach will work for most DS questions and is something you can rely on for test-day.

x^4 + y^4

1. (x - y) R 5 = 1

i.e. x - y is some number ending in 1 or 6.

There are various values of x and y that could yield such numbers, and would thus yield various different values of x^4 + y^4

INSUFFICIENT

2. (x + y) R 5 = 2

i.e. x + y is some number ending in 2 or 7.

There are various values of x and y that could yield such numbers, and would thus yield various different values of x^4 + y^4

INSUFFICIENT


Both 1 and 2:

x + y is a number ending in 2 or 7 and x - y is a number ending in 1 or 6, i.e. there are 4 combinations we need to test to see if they are possible. i.e. (x+y, x-y) are numbers ending in (2,1) (2,6) (7,1) or (7,6). It shouldn't be too tough to come up with the following examples:

19 + 13 = 32 and 19 - 13 = 6 ... satisfies (2,6)
14 + 13 = 27 and 14 - 13 = 1 ... satisfies (7,1)
19 + 18 = 37 and 19 - 18 = 1 ... satisfies (7,1)

The other 2 cases of (2,1) and (7,6) are evidently contradictory (you won't be able to find two numbers that satisfy it) and are thus not possible. Now let's test our stem for the 3 possible examples:

1. 19^4 is a number ending in 1, and 13^4 is a number ending in 1 => the sum is a number ending in 2... remainder when divided by 5 will also be 2.

2. 14^4 is a number ending in 6, and 13^4 is a number ending in 1 => the sum is a number ending in 7 ... remainder when divided by 5 will be 2.

3. 19^4 is a number ending in 1 and 18^4 is a number ending in 6 => the sum is a number ending in 7 ... remainder when divided by 5 will be 2.

Three solid cases are usually sufficient on the GMAT and this is a safe bet.

SUFFICIENT

Pick C.

Nice explanation!!

Dear AbhayPrasanna

can you please tell what calues combination we can try

btw when i tries 3 combination remainder was always 2

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

There are 2 variables (x,y) in this question, so we need 2 equations in order to match the numbers, which are provided by the 2 conditions, so there is high chance that (C) will be our answer.
Looking at the conditions, it is faster to directly substitute values into equations for question regarding remainders, so if we substitute in values (x,y)=(4,3),(9,8),(14,13)..., we will find that 4^4+3^4=256+81=337 divided by 5 gives the remainder of 2. The conditions are sufficient, so the answer is (C).

Normally for cases where we need 2 more equations, such as original conditions with 2 variable, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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