If 108 is a factor of N^{2} then which of the following could be the r

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.

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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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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Solution:

Recall that the prime factors of a perfect square all have even exponents.

Since 108 = 4 x 27 = 2^2 x 3^3, the smallest value of N^2 is 2^2 x 3^4, and hence the smallest value of N is 2 x 3^2 = 18. Of course, N also could be any multiple of 18. If N = 18k where k is odd, then the remainder when N is divided by 36 is 18. If N = 18k where k is even, then the remainder when N is divided by 36 is 0. So the remainder could be either 0 or 18.

Answer: C
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