chetan2u wrote:
gmatbusters wrote:
For how many positive integers n the expression n²+35 is a perfect square?
A) 1
B) 2
C) 3
D) 4
E) 5
GMATbuster's Collection
Let the perfect square be \(x^2\), so \(n^2+35=x^2...x^2-n^2=35...(x-n)(x+n)=35\)
Possibilities would depend on the factor of 35..
(I) \((x-n)(x+n)=35=1*35........x-n=1\) and \(x+n=35\)...Subtract the two...2n=34..n=17
(II) \((x-n)(x+n)=35=5*7........x-n=5\) and \(x+n=7\)...Subtract the two...2n=2..n=1
If we take the vice versa situation that is x-n=7 and x+n=5, we will have negative n.
So, TWO values
B
Can you explain that process a little bit closer?
I am confronted with this kind of question type for the very first time.
I had a complete wrong approach. I knew that for a number to be a perfect square it has to have a number of odd integers.
Tried to work around this frame but did not arrive at a solution.
Then I just tried to fit numbers, of course 1 made sense right from the beginning as 1*1 +35 is a perfect square.
I just don't quite get the steps involved in your approach.
You guys rewrite the equation in the form (x+n)*(x-n) = 35 and from there plug in viable numbers?
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