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
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Hello Bunuel,

Could you please explain - I am not clear on this.

The way I was approaching it was as follows:

Step 1: Find out last digit of 2 to the power 98 --> 4
Step 2: Approximate values for L and N that will give 4
Step 3: 256*L can result in numbers ending with 6/2/8/4/0, using this with values of N (0/1/2/3/4)
Step 4: Answers can be L=4 and N=0

Now here is where I get stuck - the other option is L ends in 2 and N is 2 as well. It will give 4 as the last digit.

Whats wrong with my approach? Is it even correct.

Please help!!
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If it were a four-digit number then it would be explicitly mentioned.
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I think this is a high-quality question and I agree with explanation.
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Its great question and I'd like to practice a few more of these!

Hi Brunel, so the sum or product of 2 integers is always an integer? is this the reason why you assumed \(\frac{N}{2^8}\) as integer? thanks

Hi, In the first step 2^98=2^8∗L+N

2^98 is INTEGER
2^8 is INTEGER
L IS INTEGER
So, 2^8 * L is INTEGER

Hence this is Integer = Integer + N
so N can be any integer 0,1,2,3,4

Where am i thinking wrong in this approach?

Because we are given that \(0 \le N \le 4\), no other value of N nut 0 will make N/2^28 an integer.
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