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

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Pdirienzo
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If 20/2^5 = 1/2^m + 1/2^n what is nm ? [#permalink] New post   Updated on: 19 Jun 2023, 09:30
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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
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B

Originally posted by Pdirienzo on 19 Feb 2019, 11:05.
Last edited by Bunuel on 19 Jun 2023, 09:30, edited 4 times in total.
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Re: If 20/2^5 = 1/2^m + 1/2^n what is nm ? [#permalink] New post   23 Feb 2019, 07:30
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\(\frac{20}{2^5} = \frac{1}{2^m} + \frac{1}{2^n}\)

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

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

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

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

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

thus mn = 3
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Re: If 20/2^5 = 1/2^m + 1/2^n what is nm ? [#permalink] New post   21 Feb 2019, 18:19
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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.

Answer: B
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Re: If 20/2^5 = 1/2^m + 1/2^n what is nm ? [#permalink] New post   19 Feb 2019, 14:03
Guys....anyone knows..how to solve this..appreciate responses
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Re: If 20/2^5 = 1/2^m + 1/2^n what is nm ? [#permalink] New post   Updated on: 15 Jun 2019, 12:37
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\(\frac{20}{2^5} = \frac{1}{2^m} + \frac{1}{2^n}\)
\(\frac{5}{2^3} = \frac{1}{2^m} + \frac{1}{2^n}\)
\(5 = \frac{2^3}{2^m} + \frac{2^3}{2^n}\)
\(5 = 2^{3-m} + 2^{3-n}\)

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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Untitled-1.png [ 40.45 KiB | Viewed 15879 times ]


Originally posted by MahmoudFawzy on 19 Feb 2019, 15:40.
Last edited by MahmoudFawzy on 15 Jun 2019, 12:37, edited 1 time in total.
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Re: If 20/2^5 = 1/2^m + 1/2^n what is nm ? [#permalink] New post   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 20/2^5 = 1/2^m + 1/2^n what is nm ? [#permalink] New post   20 Feb 2019, 02:22
Sorry, there was a mistake in the question, already edited the original
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Re: If 20/2^5 = 1/2^m + 1/2^n what is nm ? [#permalink] New post   Updated on: 15 Jun 2019, 12:43
Pdirienzo wrote:
Sorry, there was a mistake in the question, already edited the original

Thanks

Originally posted by MahmoudFawzy on 20 Feb 2019, 02:47.
Last edited by MahmoudFawzy on 15 Jun 2019, 12:43, edited 1 time in total.
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Re: If 20/2^5 = 1/2^m + 1/2^n what is nm ? [#permalink] New post   20 Feb 2019, 03:14
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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.

Please award kudos if helpful

Posted from my mobile device
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Re: If 20/2^5 = 1/2^m + 1/2^n what is nm ? [#permalink] New post   15 Jun 2019, 11:07
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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 20/2^5 = 1/2^m + 1/2^n what is nm ? [#permalink] New post   03 Jul 2020, 12:57
\(\frac{20}{2^5}=\frac{1}{2^m}+\frac{1}{2^n}\)
\(\frac{5}{2^3}=\frac{1}{2^m}+\frac{1}{2^n}\)
\(5=\frac{2^3}{2^m}+\frac{2^3}{2^n}\)
\(5=2^(^3^-^m^)+2^(^3^-^n^)\)

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)

\(mn=3*1=3\)
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Re: If 20/2^5 = 1/2^m + 1/2^n what is nm ? [#permalink] New post   20 Jul 2020, 04:05
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
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Re: If 20/2^5 = 1/2^m + 1/2^n what is nm ? [#permalink] New post   17 Oct 2020, 20:07
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



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
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Re: If 20/2^5 = 1/2^m + 1/2^n what is nm ? [#permalink] New post   30 Oct 2022, 23:22
ScottTargetTestPrep wrote:
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.

Answer: B


Bunuel ScottTargetTestPrep

Can you please help me understand the mistake I am making while solving this question. Please refer attachment
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Re: 20 / 2^5 = (1/ 2^m ) + ( 1 / 2^n ) What is m+n ? A) 2 b) 3 [#permalink] New post   19 Jun 2023, 09:30
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Re: 20 / 2^5 = (1/ 2^m ) + ( 1 / 2^n ) What is m+n ? A) 2 b) 3 [#permalink]
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