ans = [spoiler]3 = b[/spoiler]

20/2^5 = (1/2^m) + (1/2^n)
20 = (2^5/2^m) + (2^5/2^n)
20 = 2^(5-m) + 2^(5-n)

both terms are less than 20, therefore each term equates to one of the following:
1,2,4,8,16

Of these numbers only 4 + 16 can be added to get 20

therefore
20 = 2^(5-m) + 2^(5-n)
20 = 2^(5-1) + 2^(5-3) = 16 + 4
m=1 and n=3 , mn = 3

value of m could be 3 and n could be 1. It doesn't matter because the product gives the same answer of 3

Thanks CEO.

20/2^5 = (1/2^m)+ (1/2^n)= [(2^n) + (2^m)]/2^(m+n) ]

Thus, m+n = 5 by comparing denominator-----1
but by comparing numerators, we get n= 4 and m = 2 as 2^4 + 2^2 = 20 i.e. numerator------2
Value of m+n doesn't match from 1 and 2. What is wrong here. Tried this way in exam and got stuck there.

anks17 wrote:Thanks CEO.

20/2^5 = (1/2^m)+ (1/2^n)= [(2^n) + (2^m)]/2^(m+n) ]

Thus, m+n = 5 by comparing denominator-----1
but by comparing numerators, we get n= 4 and m = 2 as 2^4 + 2^2 = 20 i.e. numerator------2
Value of m+n doesn't match from 1 and 2. What is wrong here. Tried this way in exam and got stuck there.

Hi anks17,

The mistake is to say 2^(m+n) = 2^5 by comparing the denominator !

take a look at this example:
2/8 = 1/4
2/2^3 = 1/2^2

based on what you did you would be saying 3 = 2

If this is still not clear let me know

Yup, I assumed that if a/b=c/d then a=b, which may not be true.

Why can't you do it this way :

20/2^5 = 1/2^m + 1/2^n

Take the left side of the equation : (2^2 * 5^1)/2^5 = 5^1/2^3

Right Side = (2^m + 2^n) / 2^ m+n

Comparing the numerator of both sides, 5 = 2^m + 2^n. . This can be solved only for m = 2 and n=1 ( or m =1 and n=2)

Therefore mn = 2.

Let me put m = 2 and n=1 back in the original equation, then both side equal = 5/8