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

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What is the remainder when x^4 + y^4 divided by 5? [#permalink]  04 Jul 2009, 22:08
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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

[Reveal] Spoiler:
I got this from some other forum. Here is the link
http://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.
[Reveal] Spoiler: OA
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  05 Jul 2009, 05:29
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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.
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  06 Jul 2009, 10:52
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.
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  06 Jul 2009, 11:20
Sunchaser20 wrote:

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.

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

If anyone thinks otherwise, please elaborate.
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  06 Jul 2009, 12:27
In my approach; if it is not provided that x and y integer; the answer is E.
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  06 Jul 2009, 13:41
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.
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  07 Jul 2009, 01:40
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.
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  07 Jul 2009, 03:53
Sunchaser20 wrote:
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.

I guess this explains it quite well.
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  07 Jul 2009, 21:01
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.
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  07 Jul 2009, 22:48
Always as I am here.
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  13 Jul 2009, 11:30
Guys, I just heard about remainder theorm today can you please explain me this statement ?

Sum up A) and B):

Quote:
X-Y+X+Y=5a+1+5b+2 -> 2X=5(a+b)+3: when 2X is divided by 5, the remainder is 3.

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

Quote:
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.
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  14 Jul 2009, 05:10
skpMatcha wrote:
Guys, I just heard about remainder theorm today can you please explain me this statement ?

Sum up A) and B):

Quote:
X-Y+X+Y=5a+1+5b+2 -> 2X=5(a+b)+3: when 2X is divided by 5, the remainder is 3.

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

Quote:
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.

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
Quote:
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.

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

Last edited by Sunchaser20 on 14 Jul 2009, 11:08, edited 1 time in total.
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  14 Jul 2009, 08:07
Hey thanks for explaining, I still have few doubts..

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, 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 is divided by 5, and the answer is C.

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

Quote:
2. n is even, i.e. n=2m (m is integer), then , which means, that when X is divided by 5, the "remainder" is 3/2
skpMatcha wrote:
Guys, I just heard about remainder theorm today can you please explain me this statement ?

Sum up A) and B):

Quote:
X-Y+X+Y=5a+1+5b+2 -> 2X=5(a+b)+3: when 2X is divided by 5, the remainder is 3.

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

Quote:
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.

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 $$X=\frac{2X}{X}=\frac{5n+3}{2}=\frac{5*(2m+1)+3}{2}=\frac{5*2m+5+3}{2}=5m+\frac{8}{2}$$, which means, that when X is divided by 5, the remainder is 4, which is the case, considered by
Quote:
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.

2. n is even, i.e. n=2m (m is integer), then $$X=\frac{2X}{X}=\frac{5n+3}{2}=\frac{5*2m+3}{2}=5m+\frac{3}{2}$$, which means, that when X is divided by 5, the "remainder" is 3/2
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  14 Jul 2009, 11:08
skpMatcha wrote:
Hey thanks for explaining, I still have few doubts..

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, 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 is divided by 5, and the answer is C.

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:"
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  14 Jul 2009, 12:20
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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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  20 Jul 2010, 04:02
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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
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  22 Jul 2010, 04:10
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.
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  22 Jul 2010, 04:15
Nice explanation!!
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  13 Aug 2011, 11:35
AbhayPrasanna wrote:
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

Dear AbhayPrasanna

can you please tell what calues combination we can try

btw when i tries 3 combination remainder was always 2
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Re: What is the remainder when x^4 + y^4 divided by 5? [#permalink]  02 Feb 2015, 08:09
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