if x and y are integers

[ting.cui10]

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

OA: C
source: gmat prep

how do you solve this problem?

[nileshdalvimumbai]

x, y are Integers , Remainder of x^2 + y^2 when divided by 5 is determinable?


1. Remainder of x-y when it is divided by 5 is 1.

We cannot reach x^2 + y^2 from here unless we know the remainder when xy is divided by 5.

2. Remainder of x+y when it is divided by 5 is 2.

Same as above. Insufficient.

Combined.

x - y = 5a + 1.
x + y = 5b + 2

a and b are Integers.

Now, x ^2 + y ^2 can be found by eliminating xy from the addition of (x + y)^2 and (x - y)^2.

So, (x + y)^2 + (x - y)^2. = 2 (x^2 + y^2)

RHS = 5 [ (a^2 + b^2 + a + 2b + 1)].
The complex term involving a and b is an integer since a and b are integers.
So, crossmultiplying, we get

(x^2 + y^2)/5 = 2 * Complex Term = 2*Integer = Integer.

So the remainder is zero. C is sufficient.

[ting.cui10]

is there an non-algebraic way to solve this problem?

[RonPurewal]

hi --

what's the source of this problem?

the problem as you've written it here can't be correct:

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

notice the boldface part: with this wording, the problem is actually asking for two different remainders.
it's very important that you post problems exactly as originally written. do not change them in any way.

in any case:

* if this problem is actually from the official gmat prep software, then please post a screenshot as proof.

* if it is not, then post another thread in the General Math forum, citing the original source.

thanks.

[ting.cui10]

is there a non-algebraic way to do the problem?

[krishnan.anju1987]

The issue is to find the value of (x^2+y^2)

given (x+y)/2 gives a remainder of 2
(x-y)/2 gives a remainder of 1

now both statements are insufficient to solve the problem as they give only remainders of (x_+ y) and (x-y). Since we dont know the individual values of x and y.

Now consider the statements together.

let's take x=7, y=6

(x+y)/5=42/5 gives remainder of 2
(x-y)/5=1/5 gives a remainder of 1

7^2+6^2=49+36
sum is divisible by 5

another example

x=14, y=13
(x+y)/5=27.2 gives remainder as 2
(x-y)/5= gives one as remainder

169+196 is divisible by 5, remainder is 0

another example
x=4, y=3
x+y=7, divided by 5 gives 2 as remainder
x-y=1 divided by 5 gives one as remainder

16+9 is divisible by 5 and hence remainder is 0

hence remainder will always be 0. Hence, Sufficient.

Thanks,
Krishna

[RonPurewal]

is there a non-algebraic way to do the problem?

well, you can just plug in a whole bunch of numbers and see what happens. if you get the same result over and over, then pick "sufficient". if you get different results, pick "insufficient".

first, make lists.
* (x - y) could be 1, 6, 11, 16, 21, ...
* (x + y) could be 2, 7, 12, 17, 22, ...

now, try possibilities.
* x - y = 1, x + y = 2
doesn't work (not integers)
* x - y = 1, x + y = 7
gives x = 4, y = 3
x^2 + y^2 = 25
remainder = 0
* x - y = 1, x + y = 12
doesn't work (not integers)
* x - y = 1, x + y = 17
gives x = 9, y = 8
x^2 + y^2 = 81 + 64 = something that ends with "5"
remainder = 0
ok, that's enough of x - y = 1.
* x - y = 6, x + y = 2
gives x = 4, y = -2
x^2 + y^2 = 20
remainder = 0
* x - y = 6, x + y = 7
doesn't work (not integers)
* x - y = 6, x + y = 12
gives x = 9, y = 3
x^2 + y^2 = 90
remainder = 0

... you'll keep getting 0, over and over and over again, until you are finally convinced.

[jtl.buckeye]

Nevermind. I figured out my own mistake.

Hi,

I know the answer is C, but I just have trouble understanding why X^2+Y^2 is divisible by 5. Here's my rationale:

(x-y)/5 = p +1, where p is some integer, and 1 is the remainder
(x+y)/5 = q+2, where same as above

This translates to (x-y)= 5p+1 and (x+y)= 5q+2

x^2+y^2 = (x-y)(x+y)= (5p+1)(5q+2)=25pq+10p+5q+2--thus, the remainder would be 2, no?

I understand GMAC's explanation, but just curious why my rationale is wrong. Any help would be greatly appreciated.

My rationale is based on MGMAT's number properties guide.

[RonPurewal]

I understand GMAC's explanation, but just curious why my rationale is wrong. Any help would be greatly appreciated.

you can't factor x^2 + y^2 into (x + y)(x - y). in fact, you can't factor x^2 + y^2 into anything at all.

you are thinking of x^2 - y^2, not x^2 + y^2.

[ibiza.traffic]

Ron,

I don't understand how the cross-multiplying in the last part works....

We have 2(x^2 + y^2) = 5(int) so if we cross multiply we have
(x^2 + y^2) / 5 = (int) / 2
and we don't know if (int) is even so that the remainder will be 0..

can you please explain??


Also, the explanation on GMAT Prep says that:
2(x² + y²) = 5(5a² + 5b² + 2a + 4b + 1), which clearly implies that 2(x² + y²) is divisible by 5 with remainder = 0 and so x² + y² is also divisible by 5 with remainder = 0

I am not sure I understand this..
If 2(x^2 + y^2) is divisible by 5 why does x^2 + y^2 have to be divisible by 5 as well?
For example if we 2n is divisible by 10, n may or may not be divisible by 10 (n could be 5 or n could be 10,20 etc).

[tim]

i'm not sure what you're asking about regarding cross multiplying. your post did something very different from cross multiplying. as for why (x^2+y^2) has to be divisible by 5, it's because 5 is a prime number, so if the whole expression is divisible by 5 one part of it must be, and 2 is obviously not divisible by 5. this does not hold when you are dealing with non-prime numbers such as 10..

[ghong14]

http://postimg.org/image/44jzn1z87/

If X and Y are integers, 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


I was trying to do this problem with picking numbers, i.e. numbers that satisfy x-y /5 with a remainder of 1 and x+y/5 with a remainder of 2. I then switched over to an algebraic method:

x-y = 5a+1 a could be 6, 11, 16........(not sure how this even helps)
x+y= 5b+2 b could be 7, 12, 17........

However, I didn't know what to do after that. Not sure if the step was even necessary at all. Is there a more comprehensive approach to solving this problem?

[RonPurewal]

Since there is apparently trouble on this thread, I'll lock this thread, and leave your new one open.

See here
if-x-and-y-are-integers-what-is-the-remainder-when-x-2-y-2-t23205.html