General Math Question - Factoring

[nstannard]

In the homework we have this problem:

((8^2)(3^3)(2^4)) / (96^2)

A. 3
B. 6
C. 9
D. 12
E. 18


There has to be an easier way than squaring 96, then factoring to get the 8, 3, and 2's out.

Any thoughts?

[Ben Ku]

When working with complicated exponent problems, you should always reduce the bases to prime factors.

((8^2)(3^3)(2^4)) / (96^2)
= (((2^3)^2)(3^3)(2^4)) / (((2^5)(3^1))^2)
= ((2^6)(3^3)(2^4)) / ((2^10)(3^2))
= ((2^10)(3^3)) / ((2^10)(3^2))
= 3

Hope that helps.