I didn't understand how statement 2 can help us?
Can anybody please explain?

nisagl750 wrote:I didn't understand how statement 2 can help us?
Can anybody please explain?

If the sum of the digits of integer x is a multiple of 3, then x itself is a multiple of 3.

10^m + N implies the following:
If m=1, then 10^m + N = 10 + N = 1N.
If m=2, then 10^m + N = 100 + N = 10N.
If m=3, then 10^m + N = 1000 + N = 100N.
And so on.

In each case:
The first digit is 1, the units digit is N, and the intervening digits are all 0.
Thus, the sum of the digits = 1+N.

Statement 2: The remainder of N/3 is 2.
In other words, N is 2 more than a multiple of 3:
N = 3k + 2 = 2, 5, 8, 11, 14...
Thus, the sum of the digits of 10^m + N = 1+N = 3, 6, 9, 12, 15...
Since in each case the sum of the digits is a multiple of 3, 10^m + N is a multiple of 3.
Thus, when 10^m + N is divided by 3, the remainder is 0.
Thus, the remainder of (10^m + N)/3 cannot be greater than the remainder of (10^n + m)/3.
SUFFICIENT.

The correct answer is B.

Do we need to spend more time in putting different values for m & n to show that statement 1 is not sufficient Or there is some logic to rule out Statement 1?

P.S: Thanks for the explanations