If 2^98 = 256*m + n, where m and n are integers and

Question

If  2^{98} = 256*m + n , where  m  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

Bunuel · Accepted Answer

If  2^{98} = 256*m + n , where  m  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 * m + n . Divide both sides by  2^8 :  2^{90} = m + \frac{n}{2^8} . Since both  2^{90}  and  m  are integers, then  \frac{n}{2^8}  must also be an integer. Considering the constraint  0 \le n \le 4 , this is only possible when  n = 0 .

Answer: A

GyanOne · Answer

The question has not been stated in the correct form. Looking into the tests, I see that the actual question is:

If 2^98= 256L + N and 0<=N<=4, what is the value of N?

To solve this, note that 2^8 = 256, so 2^98 = 2^ =4*

As this is a multiple of 256, N can only be 0.

Choice A

TeamGMATIFY · Answer

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

bhavyagmat · Answer

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

bumpbot · Answer

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If 2^98 = 256*m + n, where m and n are integers and

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If 2^98 = 256*m + n, where m and n are integers and

Prep Toolkit

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