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Re: if x and y are integers [#permalink]
26 Apr 2012, 20:11

kt750 wrote:

if x and y are integer, what is the remainder when x^2 and y^2 is divided by 5?

(1) when x-y is divided by 5, the remainder is 1

(2) when x+y is divided by 5, the remainder is 2

how do you solve this problem?

My Answer is E becasue (x^2 + y^2)/5 can't be calculated either with x-y or with x+y.

Cheers. _________________

----------------------------------------------------------------------------------------- What you do TODAY is important because you're exchanging a day of your life for it! -----------------------------------------------------------------------------------------

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]
26 Apr 2012, 21:11

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If x and y are integer, what is the remainder when x^2 + y^2 is divided by 5?

(1) When x-y is divided by 5, the remainder is 1 --> x-y=5q+1, so x-y can be 1, 6, 11, ... Now, x=2 and y=1 (x-y=1) then x^2+y^2=5 and thus the remainder is 0, but if x=3 and y=2 (x-y=1) then x^2+y^2=13 and thus the remainder is 3. Not sufficient.

(2) When x+y is divided by 5, the remainder is 2 --> x+y=5p+2, so x+y can be 2, 7, 12, ... Now, x=1 and y=1 (x+y=2) then x^2+y^2=2 and thus the remainder is 2, but if x=5 and y=2 (x+y=7) then x^2+y^2=29 and thus the remainder is 4. Not sufficient.

(1)+(2) Square both expressions: x^2-2xy+y^2=25q^2+10q+1 and x^2+2xy+y^2=25p^2+20p+4 --> add them up: 2(x^2+y^2)=5(5q^2+2q+5p^2+4p+1) --> so 2(x^2+y^2) is divisible by 5 (remainder 0), which means that so is x^2+y^2. Sufficient.

Re: if x and y are integers [#permalink]
16 May 2012, 10:59

raingary wrote:

1) x-y= 5k+1 tells us only about the difference between x and y , not about the individual numbers. hence not sufficient. 2) x+y=5m+2 tells us only about the addition of x and y , not about the individual numbers. hence not sufficient. 1&2) adding both the conditions: 2x= 5k+5m+3 2x= 5(k+m)+3 x= 1/2(5(k+m)+3) inserting some positive values of k,m we can find x and thereby x^2/5.

for y subtract both the conditions: 2y= 5m+2 - 5k-1 2y= 5(m-k)+1 y= 1/2(5(m-k)+1)

inserting some positive values of k,m we can find y and thereby y^2/5.

So, C is the answer.

You've arrived at the correct answer, but your reasoning should be more on the grounds of reasoning why x-y leaving a remainder of 1 cannot provide sufficient information to ascertain if x^2+y^2 is divisible by 5.

Quick way to arrive to this in the exam is the following:

You know that 2(x^2+y^2)= (x-y)^2 + (x+y)^2. You've understood that x-y = 5q + 1 and that x+y = 5p+2. So, now we have:

2(x^2+y^2) = (5q+1)^2 + (5p+2)^2.

You know that expression is going to be divisible by 5....except for the units part. If the units part is divisible by 5, you're done. That's clearly the case just by looking at the expansion, as you will get 4+1.

So, as Bunuel put it, if twice the number is divisible by 5 (think 20), then the initial number (think 10) is also divisible by 5, and you're done.

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]
28 Aug 2012, 13:25

Question to Bunuel:

Please tell me, how in 2 minutes, looking at the question for the first time in my life, I can decifer that I shoud firstly square both equations, then ADD them and only then I will see that it can be divided by 5 too. where is the key to start such calculation? there is no time to do different approaches.

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]
29 Aug 2012, 00:17

Expert's post

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rockfeld wrote:

Question to Bunuel:

Please tell me, how in 2 minutes, looking at the question for the first time in my life, I can decifer that I shoud firstly square both equations, then ADD them and only then I will see that it can be divided by 5 too. where is the key to start such calculation? there is no time to do different approaches.

We know that x-y=5q+1 and x+y=5p+2 and we need to find the remainder when x^2 + y^2 (x squared plus y squared ) is divided by 5, so we need to get some expression, from these two, where x and y are squared and add up. Squaring and adding seems to be the best way to proceed.

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]
31 Aug 2012, 03:17

My approach, combining both statements:

(5n+r)^2=25n^2+10nr+r^2 25n^2 is divisible by 5, 10nr too. So we only have to square the remainder to find the remainder of the square of n. The same is true for adding and subtracting numbers, which should be clear. Remainder of x is 1 greater than remainder of y (because of statement 1). --> Options for remainders of x and y: (1,0), (2,1), (3,2), (4,3), (5,4)-->(0,4) Statement 2 tells us to add x+y, so remainders of our options are: 1+0=1, 2+1=3, 3+2=5+0, 4+3=5+2, 0+4=4 Statement 2 says that the remainder is 2, when adding. Only (4,3) accomplishes this. So 4^2+3^2=16+9=25 The remainder will always be 0 and both statements combined are therefore sufficient.

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]
03 Sep 2012, 13:56

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kt750 wrote:

If x and y are integer, what is the remainder when x^2 + y^2 is divided by 5?

(1) When x-y is divided by 5, the remainder is 1

(2) When x+y is divided by 5, the remainder is 2

Neither (1) nor (2) alone is sufficient.

(1) and (2) together:

From (1), x-y=5a+1 for some integer a, and from (2), x+y = 5b+2, for some integer b. (x-y)(x+y) = x^2-y^2=(5a+1)(5b+2)=M5+2 (multiple of 5 plus 2). Integers can be of the form M5, M5+1, M5-1(=M5+4), M5+2 or M5-2(=M5+3). Therefore, square of an integer can be of the form M5, M5+1 or M5-1(=M5+4).

Since x^2-y^2 is M5+2, necessarily x^2 must be M5+1 and y^2 must be M5-1. Then the sum x^2+y^2 is M5 (multiple of 5). Sufficient.

Answer C _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]
03 Sep 2012, 21:13

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kt750 wrote:

If x and y are integer, what is the remainder when x^2 + y^2 is divided by 5?

(1) When x-y is divided by 5, the remainder is 1

(2) When x+y is divided by 5, the remainder is 2

The question can be done quickly if you realize that you need to get x^2 + y^2 using the terms x-y and/or x+y. Either term alone is not sufficient to represent x^2 + y^2 but using both,

Every term in (5a + 1)^2 and (5b + 2)^2 will be divisible by 5 except last ones i.e. 1^2 and 2^2. 1+4 = 5 so the remainder when 2(x^2 + y^2) is divided by 5 is 0. So x^2 + y^2 must be a multiple of 5 and the remainder when x^2 + y^2 is divided by 5 must be 0. _________________

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]
27 Jan 2013, 20:53

Bunuel wrote:

vikram4689 wrote:

Both are correct examples but the one i gave does not follow what you mentioned

You said since 2*P is divisible by 5, therefore P is divisible by 5... This is not correct P=5/2

Not so. Your example is NOT correct.

I'm saying that if p is an integer and 2p is divisible by 5 then p must be divisible by 5. Now, check your example, is it correct?

It should be: 2*5 is divisible by 5 (p=5) so 5 (p) is divisible by 5.

Hope it's clear.

I believe vikram4689 is correct. If we know that 2P is divisible by 5, then we only know that P is divisible by 2.5.

Let, P=x^2+y^2 and N=5a^2+2a+5b^2+4b+1

Therefore, we have already proven:

2P=5N P=(5/2)N

The only way for P to be divisible by 5 is if N is even (contains a factor of 2). Unfortunately, we can't factor out a 2 from N. Therefore, whether or not N is even depends on the values of A and B.

By looking at the equation of N, we can see that it will be even only if A and B are not both positive or both negative.

It works out that the equations (x-y)=5a+1 and (x+y)=5b+2 can both be satisfied only when A and B are not both positive or both negative. This is why Bunuel's incorrect assumption still works in this case.

There is no way I could have figured all of this out in 2 or 3 minutes though.

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]
27 Jan 2013, 21:32

Expert's post

matt8 wrote:

Bunuel wrote:

vikram4689 wrote:

Both are correct examples but the one i gave does not follow what you mentioned

You said since 2*P is divisible by 5, therefore P is divisible by 5... This is not correct P=5/2

Not so. Your example is NOT correct.

I'm saying that if p is an integer and 2p is divisible by 5 then p must be divisible by 5. Now, check your example, is it correct?

It should be: 2*5 is divisible by 5 (p=5) so 5 (p) is divisible by 5.

Hope it's clear.

I believe vikram4689 is correct. If we know that 2P is divisible by 5, then we only know that P is divisible by 2.5.

Let, P=x^2+y^2 and N=5a^2+2a+5b^2+4b+1

Therefore, we have already proven:

2P=5N P=(5/2)N

The only way for P to be divisible by 5 is if N is even (contains a factor of 2). Unfortunately, we can't factor out a 2 from N. Therefore, whether or not N is even depends on the values of A and B.

By looking at the equation of N, we can see that it will be even only if A and B are not both positive or both negative.

It works out that the equations (x-y)=5a+1 and (x+y)=5b+2 can both be satisfied only when A and B are not both positive or both negative. This is why Bunuel's incorrect assumption still works in this case.

There is no way I could have figured all of this out in 2 or 3 minutes though.

Think again - It's not 'Bunuel's incorrect assumption'. It is given that x and y are integers so, given that P=x^2+y^2, P will be a positive integer. Now that we know that P is an integer, if 2P is divisible by 5, P must be divisible by 5.

Given that a and b are positive integers and that 'a' is divisible by 'b', a MUST have b as a factor. It's a basic mathematical fact. When I say, 'a' is divisible by 'b', I mean 'a' is a multiple of 'b' which means 'b' is a factor of 'a'. Try some examples to convince yourself.

This statement has far reaching implications. Think of questions like: Is 2^{100} divisible by 3? I will say 'no' without a thought. Is 2^{10}*3^{100}*5 divisible by 7? Again, absolutely no!

To reiterate, if 'a' is to be divisible by 'b', a must have b as a factor. (talking about positive integers) _________________

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]
30 Jan 2013, 21:48

Bunuel wrote:

If x and y are integer, what is the remainder when x^2 + y^2 is divided by 5?

(1) When x-y is divided by 5, the remainder is 1 --> x-y=5q+1, so x-y can be 1, 6, 11, ... Now, x=2 and y=1 (x-y=1) then x^2+y^2=5 and thus the remainder is 0, but if x=3 and y=2 (x-y=1) then x^2+y^2=13 and thus the remainder is 3. Not sufficient.

(2) When x+y is divided by 5, the remainder is 2 --> x+y=5p+2, so x+y can be 2, 7, 12, ... Now, x=1 and y=1 (x+y=2) then x^2+y^2=2 and thus the remainder is 2, but if x=5 and y=2 (x+y=7) then x^2+y^2=29 and thus the remainder is 4. Not sufficient.

(1)+(2) Square both expressions: x^2-2xy+y^2=25q^2+10q+1 and x^2+2xy+y^2=25p^2+20p+4 --> add them up: 2(x^2+y^2)=5(5q^2+2q+5p^2+4p+1) --> so 2(x^2+y^2) is divisible by 5 (remainder 0), which means that so is x^2+y^2. Sufficient.

Answer: C.

Hope it's clear.

HI... y cant we just take examples...4 and 3 (14 and 13, 24 and 23)are the only nos. which will satisfy both the condition.. Hence, by checking cyclicity for 14 and 13 even...we can say that both the statements are reqd wot say?

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]
31 Jan 2013, 11:06

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rohitgupta86 wrote:

HI... y cant we just take examples...4 and 3 (14 and 13, 24 and 23)are the only nos. which will satisfy both the condition.. Hence, by checking cyclicity for 14 and 13 even...we can say that both the statements are reqd wot say?

What about numbers such as (9 and 3) or (8 and 14) or (3 and 14) etc? By just taking numbers, can you be sure that the remainder will be 0 in each case? You will need to think of the logic behind it - you can either check for numbers and then look for logic or you can jump to the logic straight away. (Statement 1 says that the remainder when x-y is divided by 5 will be 1, not that the difference between x and y is 1)

As a general rule, it is very hard to establish that the result will be the same in every case using number plugging. It is much easier to say that it will not be the same in every case using this strategy. So number plugging is best avoided in DS until and unless you feel that taking some easy numbers will show you that the result will not be the same in every case. _________________