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If (n+12)^2 is divisible by n, where n is a natural number

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If (n+12)^2 is divisible by n, where n is a natural number  [#permalink]

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New post 29 Sep 2018, 18:15
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If \((n+12)^2\) is divisible by n, where n is a natural number. How many values can take?

a) 4
b) 6
c) 12
d) 15
e) 22

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Re: If (n+12)^2 is divisible by n, where n is a natural number  [#permalink]

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New post 29 Sep 2018, 18:43
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On expansion:- (n+12)^2 = n^2 + 24n + 144

Now [n^2 + 24n + 144] /n = n + 24 + 144/n

Here = 144 = 2^4 X 3^2
Thus 144 has (4+1)( 2+1)= 15 factors.

So n can be any one of the factors of 144 to fully divide the expression.

So n can take 15 values.
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Re: If (n+12)^2 is divisible by n, where n is a natural number  [#permalink]

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New post 29 Sep 2018, 19:14
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Afc0892 wrote:
If \((n+12)^2\) is divisible by n, where n is a natural number. How many values can take?

a) 4
b) 6
c) 12
d) 15
e) 22


\((n+12)^2\) is divisible by n or \(n^2+24n+144\) is divisible by n..
n^2+24n are clearly divisible by n, so 144 also must be divisible by n..
So n can be one of the factors of 144..
\(144=12*12=2^4*3^2\)
Number of factors = (4+1)(2+1)= 5*3 = 15

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Re: If (n+12)^2 is divisible by n, where n is a natural number   [#permalink] 29 Sep 2018, 19:14
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If (n+12)^2 is divisible by n, where n is a natural number

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