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If 2^(98)=256L+N

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manalq8
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If 2^(98)=256L+N [#permalink] New post  12 Jan 2012, 07:53
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If \(2^{98} = 256L + N\), where \(L\) and \(N\) are integers and \(0 \le N \le 4\) , what is the value of \(N\) ?

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Re: if 2^98 [#permalink] New post  12 Jan 2012, 08:29
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manalq8 wrote:
If \(2^{98} = 256L + N\) , where \(L\) and \(N\) are integers and \(0 \le N \le 4\) , what is the value of \(N\) ?

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1
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Number of approaches are possible.

Given: \(2^{98}=2^8*L+N\) --> divide both parts by \(2^8\) --> \(2^{90}=L+\frac{N}{2^8}\). Now, as both \(2^{90}\) and L are an integers then \(\frac{N}{2^8}\) must also be an integer, which is only possible for \(N=0\) (since \(0\leq{N}\leq4\)).

Answer: A.
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Re: If 2^(98)=256L+N [#permalink] New post  13 Jan 2012, 04:09
Ans should be A i.e. N = 0
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Re: If 2^(98)=256L+N [#permalink] New post  27 Feb 2014, 23:17
N = 2^98 - 256L

For N to be 0 < = N < = 4, value of L should be as high as possible

256L = 2^8 . L

When L = 90, N would be zero

Answer =A
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Re: If 2^(98)=256L+N [#permalink] New post  12 Jul 2015, 03:01
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manalq8 wrote:
If \(2^{98} = 256L + N\), where \(L\) and \(N\) are integers and \(0 \le N \le 4\) , what is the value of \(N\) ?

0
1
2
3
4


\(2^{98} = 256L + N\) --> \(2^{98}=2^{8}L+N\)
\(2^{98}=Even\), \(2^{8}L=Even\) --> \(N=Even\) --> eliminate 1, 3
If \(N=2\)-->\(2^{97}=2^{7}L+1\): Wrong
If \(N=4\)-->\(2^{96}=2^{6}L+1\): Wrong
--> N=0, answer A
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Re: If 2^(98)=256L+N [#permalink] New post  22 Oct 2022, 04:10
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Re: If 2^(98)=256L+N [#permalink]
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