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If N and x are positive integers such that \(N^N = 2^{160}\) and \(N^2 + 2^N\) is a multiple of \(2^x\), then what is the largest possible value of x?
A. 8
B. 9
C. 10
D. 11
E. 12
A. 8
B. 9
C. 10
D. 11
E. 12
07 Apr 2020, 01:01
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\(N^N = 2^{160}\)
\(N^N = (2^5)^{160/5}\)
\(N^N = (32)^{32}\)
N= 32 or \(2^5\)
\(N^2 + 2^N\) = \((2^5)^2 + 2^{32}\)
\(N^2 + 2^N\) = \((2^{10}) + 2^{32}\)
\(N^2 + 2^N\) = \(2^{10} (1+ 2^{22})\)
\(N^2 + 2^N\) = \(2^{10}\) * odd number
x=10
\(N^N = (2^5)^{160/5}\)
\(N^N = (32)^{32}\)
N= 32 or \(2^5\)
\(N^2 + 2^N\) = \((2^5)^2 + 2^{32}\)
\(N^2 + 2^N\) = \((2^{10}) + 2^{32}\)
\(N^2 + 2^N\) = \(2^{10} (1+ 2^{22})\)
\(N^2 + 2^N\) = \(2^{10}\) * odd number
x=10
Bunuel
If N and x are positive integers such that \(N^N = 2^{160}\) and \(N^2 + 2^N\) is a multiple of \(2^x\), then what is the largest possible value of x?
A. 8
B. 9
C. 10
D. 11
E. 12
A. 8
B. 9
C. 10
D. 11
E. 12
15 Jan 2021, 00:02
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Bookmarks
Bunuel
If N and x are positive integers such that \(N^N = 2^{160}\) and \(N^2 + 2^N\) is a multiple of \(2^x\), then what is the largest possible value of x?
A. 8
B. 9
C. 10
D. 11
E. 12
A. 8
B. 9
C. 10
D. 11
E. 12
\(N^N\) \(= 2^{160}\)
As N is a positive integer, it has to have a base of 2, so let \(N=2^a\)
So \((2^a)^{2^a}\) \(= 2^{160}\)
\(2^{a*2^a}\) \(= 2^{160}\)
Equating the powers
\(a*2^a=160=5*2^5\)
\(a=5\) and \(N=2^5=32\)
\(N^2\) + \(2^N\)=\(32^2+2^{32}=(2^5)^2+2^{32}=2^{10}+2^{32}=2^{10}(1+2^{22})\)
Therefore, the maximum value of x=10
C
