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.