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# What is the remainder when X^4 + Y^4 divide by 5 A) X-Y

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Manager
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What is the remainder when X^4 + Y^4 divide by 5 A) X-Y [#permalink]  04 Jul 2009, 22:08
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What is the remainder when X^4 + Y^4 divide by 5

A) X-Y divided by 5 gives remainder 1
B) X+Y divided by 5 gives remainder 2

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.
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Manager
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Re: Remainder [#permalink]  05 Jul 2009, 05:29
1
KUDOS
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.
Senior Manager
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Re: Remainder [#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: Remainder [#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.
Senior Manager
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Re: Remainder [#permalink]  06 Jul 2009, 12:27
In my approach; if it is not provided that x and y integer; the answer is E.
Manager
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Re: Remainder [#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.
Senior Manager
Joined: 23 Jun 2009
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Schools: UPenn, UMich, HKS, UCB, Chicago
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Kudos [?]: 92 [0], given: 60

Re: Remainder [#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.
Manager
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Re: Remainder [#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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Manager
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Re: Remainder [#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.
Senior Manager
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Re: Remainder [#permalink]  07 Jul 2009, 22:48
Always as I am here.
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Re: Remainder [#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: Remainder [#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.
Manager
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Re: Remainder [#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
Manager
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Re: Remainder [#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: Remainder [#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.
Re: Remainder   [#permalink] 14 Jul 2009, 12:20
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