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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If 108 is a factor of N^{2} then which of the following could be the r

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Joined: 25 Dec 2018
Posts: 143
Location: India
GMAT 1: 490 Q47 V13
GPA: 2.86
If 108 is a factor of N^{2} then which of the following could be the r  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 33% (01:57) correct 67% (02:09) wrong based on 82 sessions

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If 108 is a factor of N^{2} then which of the following could be the remainder when N is divided by 36?

I. o
II. 18
III. 6

A. I only
B. III only
C. I& II only
D. I,II& III
E. None of these
VP  D
Joined: 31 Oct 2013
Posts: 1493
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)
Re: If 108 is a factor of N^{2} then which of the following could be the r  [#permalink]

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3
2
akurathi12 wrote:
If 108 is a factor of N^{2} then which of the following could be the remainder when N is divided by 36?

I. o
II. 18
III. 6

A. I only
B. III only
C. I& II only
D. I,II& III
E. None of these

108 is a factor of$$N^2$$

Prime factorization will give us the hint about the size of N. The we will divide it by 36.

108 = 2*54 = 2*6*9 = 2*3*2*3*3.

we have :

$$2^2$$

$$3^3$$

$$N^2$$ is a perfect square . thus . there must be at lest another 3. we need even power to make a integer perfect.

$$2^2*3^4$$. we have to bring another 3 to make $$3^4$$.

Minimum value of$$N^2 = 2^2 * 3^4 = 324$$

$$18^2 = 324$$.

So, N must be 18 in this case.

18/36 = 0 + 18 . 18 is the remainder.

but we know minimum value of $$N^2$$ is 324. $$N^2$$ could be even more bigger. These extended values will be the multiple of 324 and in this case 324 will be considered as N.

324/36 = 9 + 0

Remainder is 0.

Thus, C is the correct answer.
##### General Discussion
Director  G
Joined: 09 Mar 2018
Posts: 991
Location: India
Re: If 108 is a factor of N^{2} then which of the following could be the r  [#permalink]

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2
1
akurathi12 wrote:
If 108 is a factor of N^{2} then which of the following could be the remainder when N is divided by 36?

I. 0
II. 18
III. 6

A. I only
B. III only
C. I& II only
D. I,II& III
E. None of these

108 is a factor of $$N^2$$ will mean that it will be fully consumed

108 = 2*2*3*3*3
$$N^2$$ = 2*2*3*3*3*3 *k
N = 2*3*3*k

36 = 2*2*3*3

This means that Minimum value of N can be 2*3*3 = 18
18/36, remainder = 18

Now k can take some other value as well but it will be in a pair, so that it can be utilized in $$N^2$$ = 36*9 =324
324/18, remainder = 0

C
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Manager  S
Joined: 01 Dec 2018
Posts: 82
If 108 is a factor of N^{2} then which of the following could be the r  [#permalink]

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1
akurathi12 wrote:
If 108 is a factor of N^{2} then which of the following could be the remainder when N is divided by 36?

I. o
II. 18
III. 6

A. I only
B. III only
C. I& II only
D. I,II& III
E. None of these

In simple words we need to make the value of N in a multiple of 36 . But first lets find what can be minimum value of N .

Minimum value has to be such that N SHOULD BE DIVIDED by 108 completely . 108 = 2^2 * 3^2 * 3 so for N to be N^2 it has to have one more 3 SO THAT IT CAN BE DIVIDED by 108 .

so N^2 = 2^2 * 3^2 * 3^2*K = 324K (where K is a no. which maintains the square value of N^2 Hence N becomes 18K (SQUARE ROOT OF 324K)
NOW N/36 =18K/36 gives us remiander 18 .

NOW this K has to have minimum pairs as 36 have in order to get divided by 36 . So 3^2 * 2k /2^2*3^2
WE OBSERVED THAT WE NEED ONE MORE 2 so that the N gets completely divided hence N= 3^2* 2^2 will become 36 .
and we get remainder zero . Option 6 cant be as N is minimum 18K.

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