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Bunuel
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Bunuel
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monir6000
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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The question is 2^98= 2^8*m + n - in that case

2^98,2^8*m both even - so n cant be odd - 1,3 are out

if n=4, then 2^98 = 2^8*m + 2^2
2^98= 2^2(2^6m+1)
2^96 = (2^6m+1)
LHS is even, RHS is odd - not possible
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Hi Bunuel, can you pls help validate my approach? I just want to understand if this question can be solved this way as well or if there are any flaws in my approach.
2^98 = 256*m + n
2^98 = 2^8*m + n
The units digit of 2^98 is 4, the units digit of RHS also has to be the same since they are the same number eventually.
M has to be 90 ` it cannot be 89 given that n<4 and it cannot be 91 since it would significantly exceed the number.
Hence,
2^98 = 2^98 + 0
So n=0
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