m^a * n^b = 200, all +ve integers. a+b = ? A. a b B. a

Question

m^a * n^b = 200, all +ve integers. a+b = ?

A. a > b
B. a * b =6

ventivish · Answer

The answer is D

m^a *n^b =200
Breaking 200 to its prime roots we get 
m^a *n^b =2^3 *5^2

Both statments satisfy the condition

gmatnub · Answer

B

m^a *n^b =200
Breaking 200 to its prime roots we get 
m^a *n^b =2^3 *5^2

But you can also have a situation where m^a *n^b =2^2 *50^1
so statement is not suff

statement 2
a*b = 6
a = 2, b = 3 or a = 3, b = 2
OR
a = 1, b = 6 or a = 6, b = 1 but this is not possible if you break down 200

therefore a*b = 6 is suff

wizardofwashington · Answer

Thanks, mate. That's a neat way to solve.

ventivish · Answer

Good job, needed to watch out for those kind of errors!

m^a * n^b = 200, all +ve integers. a+b = ? A. a > b B. a