If 20/2^5 = 1/2^m + 1/2^n what is nm ?

Question

If  \frac{20}{2^{5} } = \frac{1}{2^m} + \frac{1}{2^n}   what is nm ?

(A) 0
(B) 3
(C) 8
(D) 16
(E) 24

iPrasad · Accepted Answer

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

\frac{5}{2^3} = \frac{1}{2^m} + \frac{1}{2^n}

\frac{(4 +1)}{2^3} = \frac{1}{2^m} + \frac{1}{2^n}

\frac{(2^2+1)}{2^3} = \frac{1}{2^m} + \frac{1}{2^n}

\frac{1}{2^1} + \frac{1}{2^3} = \frac{1}{2^m} + \frac{1}{2^n}

this implies m = 1 when n = 3 or m = 3 when n = 1

thus mn = 3

ScottTargetTestPrep · Answer

(Note: Here we are assuming that both m and n are integers.)

Simplifying, we have:

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

5/2^3 = 1/(2^m) + 1/(2^n)

5 = (2^3)/(2^m) + (2^3)/(2^n)

5 = 2^(3 - m) + 2^(3 - n)

The only way to express 5 as the sum of two integer powers of 2 is 5 = 4 + 1 = 2^2 + 2^0. So, either of (3 - m) or (3 - n) is equal to 2 and the other one is equal to 0. If 3 - m = 2 and 3 - n = 0, then m = 1 and n = 3. In this case, we get mn = 3. If 3 - m = 0 and 3 - n = 2, we get m = 3 and n = 1 and again, mn = 3.

Answer: B

MahmoudFawzy · Answer

\frac{20}{2^5} = \frac{1}{2^m} + \frac{1}{2^n} 
 \frac{5}{2^3} = \frac{1}{2^m} + \frac{1}{2^n} 
 5 = \frac{2^3}{2^m} + \frac{2^3}{2^n} 
 5 = 2^{3-m} + 2^{3-n}

now 5 can be the sum of two pairs of integers (0,5),(1,4),(2,3)
it can't be (0,5) because there is no valid value for x when  2^x = 0 
it will be difficult to deal with (2,3) because finding x when  2^x = 3  is difficult without a calculator.
by trying (1,4) pair,
if  2^{3-m} = 1 = 2^0 , then  3-m = 0  and  m = 3 
if  2^{3-n} = 4 = 2^2 , then  3-n = 2  and  n = 1

so  m*n = 3

note that there is no one unique value for (m,n) pair (as shown in the graph), however, the line pass through the (3,1), (1,3) points.

Staphyk · Answer

20/(2^5)=1/2^m + 1/2^n
20/(2^5)=2^n+2^m/(2^m*2^n)
20(2^m*2^n)=(2^n+2^m)2^5
2^2*5(2^m*2^n)=(2^n+2^m)2^5
5(2^m*2^n)=(2^n+2^m)2^3
Now the only way the LHS will equate the RHS is when m=3 ,n=1 or when n=3 ,m=1
5(2^3*2^1)=(2^1+2^3)2^3
Now ,5*2=(2^1+2^3) 
So m*n =3 
Ans is B

Posted from my mobile device

kumarankit01 · Answer

Here are my two cents.

20/32 = 5/8

= (1+4)8 = 1/8 +4/8
 = 1/8 + 1/2

= 1/2^3 + 1/2^1

SO by comparing both side

we get m=3 and n=1

so mn = 3 
 
I could not write full soultion in this due to typing overhead.

Posted from my mobile device

learner311 · Answer

here is how I approached this question:
step 1 : express the numerator as a sum of powers of denominator (2 in this case)
 20 = 16 + 4 = 2^4+ 2^2

step 2: replace the above value in the equation
 \frac{2^4+2^2}{2^5} =\frac{1}{2^m} + \frac{1}{2^n}

this gives:
 \frac{2^4}{2^5}+\frac{2^2}{2^5} = \frac{1}{2^m} + \frac{1}{2^n} 
step 3: solve for m and n

\frac{1}{2^1}+\frac{1}{2^3} = \frac{1}{2^m} + \frac{1}{2^n}

hence, m*n = 3, Option (B)

BlueRocketAsh · Answer

\frac{20}{2^5}=\frac{1}{2^m}+\frac{1}{2^n} 
 \frac{5}{2^3}=\frac{1}{2^m}+\frac{1}{2^n} 
 5=\frac{2^3}{2^m}+\frac{2^3}{2^n} 
 5=2^(^3^-^m^)+2^(^3^-^n^)

Only way this can be is
 4+1=5  
 
Therefore (3-m)=2 and (3-n)=0 or (3-m)=0 or (3-n)=2
(m,m) = (1,3) or (3,1)

mn=3*1=3

Kritisood · Answer

try to get the LHS in the way RHS is written. we know that we need power of 2 in the denominator. so lets try to write 20 (numerator) as a sum of powers of two. note: we write as sums so that we can split the two terms and show them as the RHS

\(\frac{4+16}{2^5}\)

\(\frac{2^{2} + 2^{4}}{2^{5}}\)

\(\frac{2^2}{2^5}+\frac{2^4}{2^5}\)

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

\frac{4+16}{2^5}

\frac{2^{2} + 2^{4}}{2^{5}}

\frac{2^2}{2^5}+\frac{2^4}{2^5}

\frac{1}{2^3}+ \frac{1}{2^1} 
3*1=3

Jsound996 · Answer

We can reduce the left-hand side to  \frac{5}{2^3}  =  \frac{1}{2^m}  +  \frac{1}{2^n}

We can convert the right-hand side of the formula to  \frac{2^m + 2^n}{2^(mn)}

Look at the denominators only. MN has to equal 3

The answer is B

JoeSal · Answer

Hi, wanted to ask how you derived the answer from the second statement. My solutions is similar, but I cannot seem to arrive at the answer.

Deepthi13 · Answer

how did 5 by 2 ^3 come

Bunuel · Answer

\frac{20}{2^5}=

= \frac{2^2*5}{2^5}=

= \frac{5}{2^3}

sandyphamm · Answer

Consider m as the bigger number
 \frac{5}{2^3} = \frac{1+2^{m-n}}{2^m}  
--> m=3 and m-n = 2 --> n = 1
mn = 3

soniasw16 · Answer

Why can't I use the same method as per this question?  https://gmatclub.com/forum/for-integers-x-and-y-2-x-2-y-2-30-what-is-the-value-204326.html

MartyMurray · Answer

If  \frac{20}{2^{5} } = \frac{1}{2^m} + \frac{1}{2^n}   what is nm ?

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

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

20 = the sum of two numbers that are powers of 2

16 and 4 are two powers of 2 that add to 20.

20 = 16 + 4 = \frac {2^{5} }{2} + \frac{2^{5} }{2^3}

m = 1

n = 3

mn = 3

(A) 0
(B) 3
(C) 8
(D) 16
(E) 24

Correct answer:   B

If 20/2^5 = 1/2^m + 1/2^n what is nm ?

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If 20/2^5 = 1/2^m + 1/2^n what is nm ?

Prep Toolkit

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