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

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If 20/5 = 1/2^m + 1/2^n what is nm ?  [#permalink]

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Updated on: 20 Feb 2019, 02:21
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Difficulty:

65% (hard)

Question Stats:

58% (02:10) correct 42% (02:36) wrong based on 170 sessions

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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

Originally posted by Pdirienzo on 19 Feb 2019, 11:05.
Last edited by Pdirienzo on 20 Feb 2019, 02:21, edited 1 time in total.
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Re: If 20/5 = 1/2^m + 1/2^n what is nm ?  [#permalink]

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20 Feb 2019, 03:14
5
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.

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Re: If 20/5 = 1/2^m + 1/2^n what is nm ?  [#permalink]

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19 Feb 2019, 14:03
Guys....anyone knows..how to solve this..appreciate responses
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If 20/5 = 1/2^m + 1/2^n what is nm ?  [#permalink]

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Updated on: 15 Jun 2019, 12:37
1
$$\frac{20}{2^5} = \frac{1}{2^m} + \frac{1}{2^n}$$
$$\frac{5}{2^3} = \frac{1}{2^m} + \frac{1}{2^n}$$
$$5 = \frac{2^3}{2^m} + \frac{2^3}{2^n}$$
$$5 = 2^{3-m} + 2^{3-n}$$

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.
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Originally posted by Mahmoudfawzy83 on 19 Feb 2019, 15:40.
Last edited by Mahmoudfawzy83 on 15 Jun 2019, 12:37, edited 1 time in total.
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If 20/5 = 1/2^m + 1/2^n what is n?  [#permalink]

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Updated on: 20 Feb 2019, 02:47
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

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Originally posted by Staphyk on 20 Feb 2019, 00:36.
Last edited by Staphyk on 20 Feb 2019, 02:47, edited 1 time in total.
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Re: If 20/5 = 1/2^m + 1/2^n what is nm ?  [#permalink]

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20 Feb 2019, 02:22
Sorry, there was a mistake in the question, already edited the original
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If 20/5 = 1/2^m + 1/2^n what is nm ?  [#permalink]

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Updated on: 15 Jun 2019, 12:43
Pdirienzo wrote:
Sorry, there was a mistake in the question, already edited the original

Thanks
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Originally posted by Mahmoudfawzy83 on 20 Feb 2019, 02:47.
Last edited by Mahmoudfawzy83 on 15 Jun 2019, 12:43, edited 1 time in total.
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Re: If 20/5 = 1/2^m + 1/2^n what is nm ?  [#permalink]

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20 Feb 2019, 03:34
20/2^5=2^m+2^n/2^mn
4*5/2^5=2^m+2^n/2^mn
5/2^3=2^m+2^n/2^mn
so all individual values of m and n, mn should always be 3 to satisfy the equation.

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Re: If 20/5 = 1/2^m + 1/2^n what is nm ?  [#permalink]

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21 Feb 2019, 18:19
3
Pdirienzo wrote:
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

(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.

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If 20/5 = 1/2^m + 1/2^n what is nm ?  [#permalink]

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23 Feb 2019, 07:30
4
$$\frac{20}{2^5} = \frac{1}{2^m} + \frac{1}{2^n}$$

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

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

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

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

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

thus mn = 3
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Re: If 20/5 = 1/2^m + 1/2^n what is nm ?  [#permalink]

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15 Jun 2019, 11:07
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)
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Re: If 20/5 = 1/2^m + 1/2^n what is nm ?   [#permalink] 15 Jun 2019, 11:07
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