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If 20/5 = 1/2^m + 1/2^n what is nm ?
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Updated on: 20 Feb 2019, 02:21
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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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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 ?
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20 Feb 2019, 03:14
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 ?
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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 ?
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Updated on: 15 Jun 2019, 12:37
\(\frac{20}{2^5} = \frac{1}{2^m} + \frac{1}{2^n}\) \(\frac{5}{2^3} = \frac{1}{2^m} + \frac{1}{2^n}\) \(5 = \frac{2^3}{2^m} + \frac{2^3}{2^n}\) \(5 = 2^{3m} + 2^{3n}\)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 \(2^{3m} = 1 = 2^0\), then \(3m = 0\) and \(m = 3\) if \(2^{3n} = 4 = 2^2\), then \(3n = 2\) 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 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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If 20/5 = 1/2^m + 1/2^n what is n?
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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 ?
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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 ?
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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
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 20/5 = 1/2^m + 1/2^n what is nm ?
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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 ?
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21 Feb 2019, 18:19
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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If 20/5 = 1/2^m + 1/2^n what is nm ?
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23 Feb 2019, 07:30
\(\frac{20}{2^5} = \frac{1}{2^m} + \frac{1}{2^n}\)
\(\frac{5}{2^3} = \frac{1}{2^m} + \frac{1}{2^n}\)
\(\frac{(4 +1)}{2^3} = \frac{1}{2^m} + \frac{1}{2^n}\)
\(\frac{(2^2+1)}{2^3} = \frac{1}{2^m} + \frac{1}{2^n}\)
\(\frac{1}{2^1} + \frac{1}{2^3} = \frac{1}{2^m} + \frac{1}{2^n}\)
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 ?
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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 ?
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15 Jun 2019, 11:07






