Bunuel wrote:

Official Solution:

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

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 \(0 \le N \le 4\)).

Answer: A

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!!

_________________

Thanks,

aimtoteach

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