DS | Remainder when (x^2 + y^2) divided by 5 ?

[yourshail123]

If x and y are integers, what is the remainder when (x^2 + y^2) divided by 5?
A) When (x - y) is divided by 5, the remainder is 1.
B) When (x + y) is divided by 5, the remainder is 2.

[eaakbari]

IMO [spoiler]C
[/spoiler]
Info :
x^2 + y^2 = 5q + r

On a quick glance at the Statement 1 & 2, we rewrite the info as

1/2 * ((x + y)^2 + (x - y)^2) = 5q + r

(1) x-y = 5k + 1
Clearly Insuff

(2) x+y = 5m + 2
Clearly Insuff


(1) & (2) together

(x + y)^2 + (x - y)^2 = 2 (x^2 + y^2)
Putting statement (1) &(2) and simplifying

2 (x^2 + y^2) = 25k^2 + 25m^2 + 10k + 20m + 5
2 (x^2 + y^2) = 5( 5k^2 + 5m^2 + 2k + 4m + 1)

Cross-multiplying
(x^2 + y^2) / 5 = 2 * (Integer term)

Hence no remainder

Answer is C

[yourshail123]

My confusion lies here -

2 (x^2 + y^2) = 5( 5k^2 + 5m^2 + 2k + 4m + 1)
Cross-multiplying
(x^2 + y^2) / 5 = 2 * (Integer term)

After cross-multiplication: Integer term becomes ( 5k^2 + 5m^2 + 2k + 4m + 1) / 2, which may or may not be integer and so the remainder will vary.
Have you made mistake here after cross-multiplication or I am mistaken?

[eaakbari]

Cross-multiplying
(x^2 + y^2) / 5 = 2 * (Integer term)

This is my typo and careless error but nevertheless, apply the following logic

2 (x^2 + y^2) = 5( 5k^2 + 5m^2 + 2k + 4m + 1)

If 2 (x^2 + y^2) is divisible by 5, then so is (x^2 + y^2)

Hence no remainder and answer C

[GMATGuruNY]

If x and y are integers, what is the remainder when (x^2 + y^2) divided by 5?
A) When (x - y) is divided by 5, the remainder is 1.
B) When (x + y) is divided by 5, the remainder is 2.

Statement 1:
x-y = 5a + 1 = 1, 6, 11, 36...
Since x and y can take on many different integer values, INSUFFICIENT.

Statement 2:
x+y = 5b + 2 = 2, 7, 12, 17...
Since x and y can take on many different integer values, INSUFFICIENT.

Statements combined:
x-y = 1, 6, 11, 21...
x+y = 2, 7, 12, 17...
(x-y) + (x+y) = 2x.
Since x is an integer, (x-y) + (x+y) must be even.
Thus, x-y and x+y are either BOTH EVEN or BOTH ODD.

Note the following:
(x-y)² + (x+y)² = (x² - 2xy + y²) + (x² + 2xy + y²) = 2(x² + y²).

(x-y)² = 1, 36, 121, 256...
(x+y)² = 4, 49, 144, 289...

If (x-y)² and (x+y)² are BOTH EVEN, their sum will have a units digit of 0:
36+4=40.
256+144=400.
A units digit of 0 implies a multiple of 10.

If (x-y)² and (x+y)² are BOTH ODD, their sum will have a units digit of 0:
1+49=50.
121+289=410.
A units digit of 0 implies a multiple of 10.

Thus, whether (x-y)² and (x+y)² are BOTH EVEN or BOTH ODD:
(x-y)² + (x+y)² = multiple of 10.
2(x² + y²) = multiple of 10
x² + y² = (multiple of 10)/2 = multiple of 5.

Thus, when x²+y² is divided by 5, the remainder will be 0.
SUFFICIENT.

The correct answer is C.

[eaakbari]

Yourshail123,

2 (x^2 + y^2) = 25k^2 + 25m^2 + 10k + 20m + 5

2 (x^2 + y^2) = 5( 5k^2 + 5m^2 + 2k + 4m + 1)

2 *(...) = 5 *(----)

Hence 2 & 5 are prime and have no common factors, therefore (...) must be a multiple of 5 and (----)must be a multiple of 2.

(...) represents (x^2 + y^2) hence (x^2 + y^2) is a multiple of 5 and is divisible by it

HTH
[/u]

[mariofelixpasku]

is there any shorter solution? it seems as if one would need more than 2 minutes to figure all this out ...

[Anju@Gurome]

is there any shorter solution? it seems as if one would need more than 2 minutes to figure all this out ...

You can avoid a lot of algebraic calculation and number manipulations if you approach this analytically. And remember this for future reference,

• If N leaves a remainder of r when divided by d, N² will leave a remainder of p when divided by d, where p is the remainder when r² is divided by r.
  
  For example, 5 leaves a remainder of 2 when divided by 3.
  5² = 25 leaves a remainder of 1 when divided by 3 ---> 1 is the remainder of 2² when divided by 3.

In this case, it is evident that none of the statements are individually sufficient. You can pick some numbers and confirm that.

1 & 2 Together: We know that, (x - y)² + (x + y)² = 2(x² + y²)
Now, from statement 1, (x - y) leaves a remainder of 1 when divided by 5.
Hence, (x - y)² will leave a remainder of 1² = 1 when divided by 5.
And, from statement 2, (x + y) leaves a remainder of 2 when divided by 5.
Hence, (x + y)² will leave a remainder of 2² = 4 when divided by 5.

Hence, 2(x² + y²) will leave a remainder of (1 + 4) = 5 when divided by 5.
So, 2(x² + y²) is divisible by 5.
--> (x² + y²) is divisible by 5.

Sufficient

The correct answer is C.

[varun289]

Here two approaches discussed are excellent but GMAT is exam for managerial skills not for MATH + English Expertise

so i would like to suggest simple approach -

take x=2 y=1 for fact one -
take x=4, y-3 for fact two also take x=2 y=0 for fact two

now combine x=4 & y=3 will get C
as we get ONE PAIR that satisfy C

eliminate E
Take SMALL number to test fast