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B
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Difficulty:
65%
(hard)
Question Stats:
65% (02:16) correct
35%
(02:37)
wrong
based on 2268
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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
(A) 0
(B) 3
(C) 8
(D) 16
(E) 24
iPrasad
Joined: 11 May 2018
Last visit: 30 Jun 2022
Posts: 25
Given Kudos: 132
Location: India
Schools: IE Jan21 Intake (A) Schulich(Jan) '22 (A)
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
\(\frac{5}{2^3} = \frac{1}{2^m} + \frac{1}{2^n}\)
\(\frac{(4 +1)}{2^3} = \frac{1}{2^m} + \frac{1}{2^n}\)
\(\frac{(2^2+1)}{2^3} = \frac{1}{2^m} + \frac{1}{2^n}\)
\(\frac{1}{2^1} + \frac{1}{2^3} = \frac{1}{2^m} + \frac{1}{2^n}\)
this implies m = 1 when n = 3 or m = 3 when n = 1
thus mn = 3
21 Feb 2019, 18:19
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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
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
