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NUS School Moderator V
Joined: 18 Jul 2018
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If (n+12)^2 is divisible by n, where n is a natural number  [#permalink]

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4 00:00

Difficulty:   55% (hard)

Question Stats: 55% (01:47) correct 45% (01:58) wrong based on 62 sessions

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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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Current Student B
Joined: 22 Jun 2018
Posts: 28
Location: India
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Re: If (n+12)^2 is divisible by n, where n is a natural number  [#permalink]

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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.
Math Expert V
Joined: 02 Aug 2009
Posts: 7957
Re: If (n+12)^2 is divisible by n, where n is a natural number  [#permalink]

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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]

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Chethan92 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

Seems as though there is an n missing from the question?
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- Stne Re: If (n+12)^2 is divisible by n, where n is a natural number   [#permalink] 13 May 2019, 02:05
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If (n+12)^2 is divisible by n, where n is a natural number

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