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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72% (01:08) correct 
