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Question Stats:
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27%
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If \(2^{98} = 256*m + n\), where \(m\) and \(n\) are integers and \(0 \le n \le 4\), what is the value of \(n\)?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
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08 May 2012, 00:02
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If \(2^{98} = 256*m + n\), where \(m\) and \(n\) are integers and \(0 \le n \le 4\), what is the value of \(n\)?
A. 0
B. 1
C. 2
D. 3
E. 4
Given: \(2^{98} = 2^8 * m + n\). Divide both sides by \(2^8\): \(2^{90} = m + \frac{n}{2^8}\). Since both \(2^{90}\) and \(m\) are integers, then \(\frac{n}{2^8}\) must also be an integer. Considering the constraint \(0 \le n \le 4\), this is only possible when \(n = 0\).
Answer: A
A. 0
B. 1
C. 2
D. 3
E. 4
Given: \(2^{98} = 2^8 * m + n\). Divide both sides by \(2^8\): \(2^{90} = m + \frac{n}{2^8}\). Since both \(2^{90}\) and \(m\) are integers, then \(\frac{n}{2^8}\) must also be an integer. Considering the constraint \(0 \le n \le 4\), this is only possible when \(n = 0\).
Answer: A
General Discussion
07 May 2012, 21:28
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Quote:
The question has not been stated in the correct form. Looking into the tests, I see that the actual question is:
If 2^98= 256L + N and 0<=N<=4, what is the value of N?
To solve this, note that 2^8 = 256, so 2^98 = 2^[(8*12)+2]=4*[256*256*......12 times]
As this is a multiple of 256, N can only be 0.
Choice A
