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If 2^98 = 256*m + n, where m and n are integers and

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monir6000
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If 2^98 = 256*m + n, where m and n are integers and [#permalink] New post   Updated on: 26 Oct 2023, 00:36
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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

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Originally posted by monir6000 on 07 May 2012, 20:57.
Last edited by Bunuel on 26 Oct 2023, 00:36, edited 2 times in total.
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If 2^98 = 256*m + n, where m and n are integers and [#permalink] New post   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
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Re: If 2^98 = 256*m + n, where m and n are integers and [#permalink] New post   07 May 2012, 21:28
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Quote:
If 2^98+L, where L and N are integers and 0<=N<=4, what is the value of N?



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
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Re: If 2^98 = 256*m + n, where m and n are integers and [#permalink] New post   26 Nov 2015, 09:13
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monir6000 wrote:
If \(2^{98} = 256L + N\), where \(L\) 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


The first thing to note here is that LHS is an even number, hence RHS should also be even.
This elements N = 1,3

Now,
\(2^{98} = 2^8L + N\), Diving both the sides by \(2^8\)

\(2^{90} = L + N/2^8\),
Out of 0, 2 and 4, N can only be 0 for RHS to be an even number.

Option A
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Re: If 2^98 = 256*m + n, where m and n are integers and [#permalink] New post   25 Oct 2023, 13:39
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Re: If 2^98 = 256*m + n, where m and n are integers and [#permalink] New post   26 Oct 2023, 00:32
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Re: If 2^98 = 256*m + n, where m and n are integers and [#permalink]
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