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

We have: 2^{90} = m + \frac{n}{2^8} . Since 2^90 and m are integers, n/2^8 must also be an integer. But n is restricted to 0, 1, 2, 3, 4. None of those divided by 2^8 gives an integer, except 0. So the only possible value for n is 0.

I like the solution - it’s helpful.

2^98=(2^8)*2^12(expected m value as it is an integer) +n =>2^98=2^96 + n This is only possible when n=4. please clarify what's wrong in this.

This doesn’t make sense. First, assuming m = 2^12 is incorrect. In fact, solving gives m = 2^90. Next, 2^8 * 2^12 = 2^20, not 2^98. Finally, if 2^98 = 2^96 + n, then n = 237684487542793012780631851008 = 2^96*3, not 4. Exponents and Roots of Numbers Theory: Exponents and Roots on the GMAT: Tips and hints Cyclicity and the remainders on the GMAT Questions Units digits, exponents, remainders problems Tough and tricky DS exponents and roots questions with detailed solutions Tough and tricky PS exponents and roots questions with detailed solutions Exponents PS Questions Roots PS Questions Check below for more: ALL YOU NEED FOR QUANT ! ! ! Ultimate GMAT Quantitative Megathread Hope this helps.

This is a great question that’s helpful for learning. This was a pure conceptual question and threw me off for sometime but eventually I got it right.

I did not quite understand the solution. 0/2^8 = 0, as 0 divided by any number is zero (not including 0/0), and 0 is an integer. It satisfies.

The questions already species the range for n as greater than equal to 0 and less than equal to 4 If you have n/2^8 then you try to put integer values of n in this range and you will see you won’t get an integer value in return as let’s say n =2 so, 2/2^8 won’t give an integer answer. Only n=0 fits in this case as 0 is itself an integer and 0/2^8 gives 0.

Exactly, that’s the key point. Since 2^90 and m are both integers, n/2^8 must also be an integer. But n is restricted to 0 ≤ n ≤ 4, and the only value in that range that makes (n/2^8) an integer is n = 0, because 1/256, 2/256, 3/256, and 4/256 are not integers. Hence n = 0.