I approched it like this: \(\frac{1}{?} + \frac{1}{?} = \frac{5}{8}\)

\(\frac{20}{2^5} = \frac{20}{32} = \frac{5}{8}\)

When adding fractions, the denominator needs to be the same, so we need to figure up all combinations that add up to 5 for the numerator

1+4

2+3

(the inverse of each doesn't matter as it will be the same fractions adding together in a different order).

So

\(\frac{1}{8} + \frac{4}{8} = 5/8\) AND \(\frac{2}{8} + \frac{3}{8} = 5/8\)

We have to determine which of these fractions will allow us to reduce down to each having 1 as a numerator. That is \(\frac{1}{8} + \frac{4}{8} = 5/8\) becomes \(\frac{1}{8} + \frac{1}{2} = 5/8\).

Now we have to break the numerator down into base of 2 with exponents.

\(\frac{1}{2^1} + \frac{1}{2^3} = \frac{5}{8}\)

This gives us 1 and 3 so 1 + 3 for m + n = 4, or C

Much of this should be done in your head rather than on paper. THe explanation is long just to make sure everyone understands it. Speed it up for the actual test or you'll be too far behind!!

bhatiagp wrote:

\(\frac{20}{2^5} = \frac{1}{2^m} + \frac{1}{2^n}\)

What is m+n ?

A) 2

b) 3

c)4

d) 5

e) Cannot be determined

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J Allen Morris

**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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