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If 2^(98)=256L+N

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If 2^(98)=256L+N [#permalink] New post 12 Jan 2012, 06:53
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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, 07:29
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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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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, 03:09
Ans should be A i.e. N = 0
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Re: If 2^(98)=256L+N [#permalink] New post 27 Feb 2014, 22:17
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] 27 Feb 2014, 22:17
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