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# For how many positive integers n the expression n²+35

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For how many positive integers n the expression n²+35  [#permalink]

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20 Oct 2019, 23:10
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85% (hard)

Question Stats:

24% (01:25) correct 76% (01:42) wrong based on 51 sessions

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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

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Posts: 8182
Re: For how many positive integers n the expression n²+35  [#permalink]

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20 Oct 2019, 23:17
1
2
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
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For how many positive integers n the expression n²+35  [#permalink]

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20 Oct 2019, 23:22
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

Given,

$$n^2$$ + 35 = $$x^2$$

$$x^2 - n^2 = 35$$

(x + n) ( x - n) = 35

7 * 5 = 35 .................................... case 1

35*1 = 35......................................case 2

Case 1:

x + n = 6 + 1 .

x - n = 6 - 1.

n = 1 here.

Case 2:

x + n = 35

18 + 17 = 35

x - n = 1

18 - 17 = 1.

n = 17 here

Thus, n = 1, 17.

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For how many positive integers n the expression n²+35  [#permalink]

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21 Oct 2019, 07:47
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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Re: For how many positive integers n the expression n²+35  [#permalink]

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21 Oct 2019, 08:10
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

$$n^2+35 = s^2$$

Or, $$s^2 - n^2 = 35$$

Or, $$(s + n)(s - n) = 7*5$$

Now, Solve for n and S , $$s = 6$$ & $$n = 1$$

Thus, There are 2 possible values, Answer must be (B)
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Re: For how many positive integers n the expression n²+35  [#permalink]

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21 Oct 2019, 19:35
Hi chrtpmdr

After getting (x+n)*(x-n) = 35
take the possible mutiplication pairs of 35
35 = 35*1 or 7*5
now, equating (x+n)*(x-n) = 35*1
we get x+n = 35 and x-n = 1
solve these two by adding the equations, we get 2x= 36 or x = 18. hence n = 17

similarly, equating (x+n)*(x-n) = 7*5
we get x+n = 7 and x-n = 5
solve these two by adding the equations, we get 2x= 12 or x = 6. hence n = 1

hence there can be 2 positive integers n = 17 or 1 which satisfy the given condition.

chrtpmdr wrote:
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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Re: For how many positive integers n the expression n²+35  [#permalink]

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23 Oct 2019, 19:09
1
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

We let x^2 be the perfect square. So we have:

n^2 + 35 = x^2

35 = x^2 - n^2

35 = (x - n)(x + n)

We see that we have two integers (x - n) and (x + n) whose product must be 35. Thus, by inspection, we see that we can have a product of 35 by either 5 * 7 or by 1 * 35.

For the first case (5 * 7), we have:

x - n = 5

x = n + 5

and

x + n = 7

x = -n + 7

Setting the two equations equal to each other, we have:

n + 5 = -n + 7

2n = 2

n = 1

For the second case (1 * 35), we have:

x - n = 1

x = n + 1

and

x + n = 35

x = -n + 35

Setting the two equations equal to each other, we have:

n + 1 = -n + 35

2n = 34

n = 17

So, there are two cases: n = 1 and n = 17, such that n^2 + 35 is a perfect square.

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Re: For how many positive integers n the expression n²+35   [#permalink] 23 Oct 2019, 19:09
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