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 = 1
N.
If m=2, then 10^m +
N = 100 +
N = 10
N.
If m=3, then 10^m +
N = 1000 +
N = 100
N.
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.
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