Do the prime factorization. So here only 6^2 can be factorize to 2^2 and 3^2.
Ans if 5^2 and 3^3 is factor of given expression then n must have 5^2 * 3^1 = 25 * 3 = 75

So D.

Thanks for the response Abhishek!

Can you help me understand how you got to n must have a factor of 5^2 and 3^1?

5^2 as a factor is given, is the 3^1 resulting from factoring out 3^2 from 6^2? I think I'm not seeing the relationship here.

Thanks!

Buix0065 wrote:Thanks for the response Abhishek!

Can you help me understand how you got to n must have a factor of 5^2 and 3^1?

5^2 as a factor is given, is the 3^1 resulting from factoring out 3^2 from 6^2? I think I'm not seeing the relationship here.

Thanks!

For 5^2 and 3^3 to be factor of n x 2^5 x 6^2 x 7^3 you need 5^2 and 3^3 in the expression.
Now if u prim factorize and see how many 5 and 3 are there you will come to know how many more you need.

n x 2^5 x 6^2 x 7^3 = n x 2^5 x (2^2 x 3^2) x 7^3 = n x 2^7 x 3^2 x 7^3
Now you need two 5 and 1 more 3. so n must be 5^2 * 3^1.


Hope this hepls.

having a number as a factor also means it could be evenly divided out therefore the statment says:

n x 2^5 x 6^2 x 7^3/(5^2)(3^3) = an integer.

then 6 is not a prime number and so should be factored: 6^2 is the same as 6 x 6 which is the same as
(2x3)(2x3) which is the same as 2^2 x 3^2. put that into the orignal formula:

n x 2^5 x 2^2 x 3^2 x 7^3/ 5^2 3^3 = when we do the division we can cancel (factor out) a 3^2 from the top and bottom. to get

n x 2^5 x 2^2 x 7^3/ 5^2 x 3 - thus in order to factor out the rest of the denominator the n must contain a 5^2 and a 3 - therefore the least possible value of n is 25(3) or 75.

I really pulled this one apart to show technique, once you have mastered the idea of factoring it will come much quicker.

Best of Luck