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# 20 / 2^5 = (1/ 2^m ) + ( 1 / 2^n ) What is m+n ? A) 2 b) 3

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20 / 2^5 = (1/ 2^m ) + ( 1 / 2^n ) What is m+n ? A) 2 b) 3 [#permalink]  22 Jun 2008, 22:53
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20 /$$2^5$$ = (1/$$2^m$$) + ( 1 /$$2^n$$)

What is m+n ?

A) 2
b) 3
c)4
d) 5
e) Cannot be determined
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Re: Exponents [#permalink]  22 Jun 2008, 23:10
20/2^5 = 1/2^m + 1/2^n

or can be written as 5/2^3 = 1/2^m + 1/2^n
now 5 can be broken into 1+4 (other ways are not possible - 2+3 or 5+1 does not allow this eq.)

Hence m+n = 3+1 = 4
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Re: Exponents [#permalink]  23 Jun 2008, 00:11
I think the answer should be (E) - cannot be determined.

If I take any of the values in the other four options and try to satisfy the equation, it does not hold true.
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Re: Exponents [#permalink]  23 Jun 2008, 00:15
bhatiagp wrote:
20 /$$2^5$$ = (1/$$2^m$$) + ( 1 /$$2^n$$)

What is m+n ?

A) 2
b) 3
c)4
d) 5
e) Cannot be determined

20 / (2^5)
= 5/8
= 1/2 + 1/8
= 1/(2^1) + 1/(2^3)
=> m=1 & n=3

Hence (C)
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Re: Exponents [#permalink]  23 Jun 2008, 00:17
Choose E.

20/2^5 = 13/2^5+7/2^5=5/2^3 = 1/2^3+1/2^1=1/2+3/2^3 = etc
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Re: Exponents [#permalink]  23 Jun 2008, 00:42
Thanks .. The Answer is C= 4
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Re: Exponents [#permalink]  23 Jun 2008, 01:58
CAN YOU POST THE OE? AS I MENTIONED, WE CAN EXTRACT INTIMAL FRACTION INTO ANY OTHER FRACTIONS.
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Re: Exponents [#permalink]  23 Jun 2008, 02:22
20/2^5=1/2^m + 1/2^n
now 20=4*5 can be written as 2^2*5
Thus 5/2^3
again 5 can be written as 4+1=2^2+1
Thus 2^2/2^3 +1/2^3=1/2 +1/2^3
Comparing m=1 and n=3
Thus m+n=1+3=4
Hence C
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Re: Exponents [#permalink]  23 Jun 2008, 08:49
ritula wrote:
20/2^5=1/2^m + 1/2^n
now 20=4*5 can be written as 2^2*5
Thus 5/2^3
again 5 can be written as 4+1=2^2+1
Thus 2^2/2^3 +1/2^3=1/2 +1/2^3
Comparing m=1 and n=3
Thus m+n=1+3=4
Hence C

4+1=2^2+1 = 2+3
Thus 2/2^3 +3/2^3=1/2^2 +3/2^3 ==> m+n#4
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Re: Exponents [#permalink]  23 Jun 2008, 09:04
lexis wrote:
4+1=2^2+1 = 2+3
Thus 2/2^3 +3/2^3=1/2^2 +3/2^3 ==> m+n#4

You don't satisfy here the form given in the question, which is $$\frac{1}{2^m}+\frac{1}{2^n}$$

(it is ONE over 2^n, not THREE over 2^n)
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Re: Exponents [#permalink]  23 Jun 2008, 09:32
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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Last edited by jallenmorris on 23 Jun 2008, 09:36, edited 2 times in total.
Re: Exponents   [#permalink] 23 Jun 2008, 09:32
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