For the first one

Given

5^2 and

3^3 are factors of

n\times 2^5\times6^2\times7^3Need to find the smallest value of n.

As

5^2 and

3^3 are factors of

n\times 2^5\times6^2\times7^3,

n\times 2^5\times6^2\times7^3 should be either the Least common Multiple of the two or a multiple of the LCM itself. i.e. dividing

n\times 2^5\times6^2\times7^3 by the LCM should result into an integer.

Lets find the LCM of the two:

5^2=5\times5 3^3=3\times3\times3\text{LCM}=5\times5\times3\times3\times3=5^2\times3^3Now

\frac{n\times 2^5\times6^2\times7^3}{5^2\times3^3}should be an integer

=\frac{n\times 2^5\times2^2\times3^2\times7^3}{5^2\times3^3}Reducing it further

= \frac{n\times 2^5\times2^2\times7^3}{5^2\times3}for the fraction to be an integer n should be divisible by

{5^2\times3}. Ans the smallest value n can have is

{5^2\times3}.

As

\frac{5^2\times3}{5^2\times3}=1{5^2\times3}=75Hence the answer is D.

Also here is a good overview of the basics of factors.

http://www.math.com/school/subject1/les ... 3L1GL.html