If x and y are positive integers such that the square of x

Question

If x and y are positive integers such that the square of x is 315 greater than the square of y, what is the value of x?

(1) |x-20|<5
(2) x-y is an odd number

CookieMonster · Answer

is it E?


x^2 = y^2 + 315

1) 15<x<25
 x^2 - 315 = integer^2
 x can be 18 and 22 for example

2) we already know this

 x^2 - y^2 = 315
(x-y)(x+y) = 315
315 has no even factors..

nervousgmat · Answer

I am getting an E here too... What is the OA?

kwam · Answer

(E)

From 1 we know that 15<x<25
And with the original equation (X^2=Y^2+315) we have two possibilities:
X=18, Y=9
X=22, Y=13
INSUF

2)alone is insuff

Both together are still INSUFF, since the 1 already limits for a X-Y as an odd number

kidstyx · Answer

how does one get: X=18, Y=9 X=22, Y=13 just by observing that x lies between 15 and 25? is it by trial and error? :shock:

kyatin · Answer

I say (E)

(X+Y)(X-Y) = 315

315 = 5*3*3*7

so (X+Y)(X-Y) has 4 options

5*63
15*21
9*35
45*7

From above 4 combinations if we use (X+Y) = 63 (X-Y)=5 etc... 
possible XY values are

36,27
18,3
22,13
26,19

Of above, all are possible combinations except first.

Now 
1) eliminated pair 4 but leaves out 2,3
So INSUFF

2) is redundant as for 315 to be difference, (x-y) has to be ODD.

SO answer is E.

kyatin · Answer

What I dint like about this problem was after lot of efforts we got to know it was (E)

Is there a faster approach for this?

If x and y are positive integers such that the square of x

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