What is the remainder when 3 digit number ABC where A, B, and C are its hundreds, tens and units digit respectively is divided by 3?

1. A+B+C = 3K+13 where K is an integer.
2. A+B+C = 19.

The OA is D.

By the divisibility rule of 3, the sum of the digits of the number has to be divisible by 3 , to be divisible by 3.

And hence if it is, the remainder will be 0.

Let's test our pre-thinking:-

24/3 now 2+4 = 6 100% divisible hence remainder 0

236 now 2+3+6 = 11 - 11/3 remainder 2 if we divide 236/3 the remainder is actually 2.

So, we have established a pattern

Let's Examine the facts:

1. A+B+C = 3K+13

Let's test k for 1
3*1+13 = 16 - 16/3 remainder is 1 hence if we divide ABC by 3 the remainder will be 1.

If k = -1
3*-1 +13 = 10 - 10/3 - remainder 1.

If K=0
3*0+13 =13 - 13/3 remainder 1.

Hence Fact 1 is sufficient.

Fact 2

A+B+C = 19 - 19/3 remainder 1.
Sufficient.

Hence the answer is D.

Has anyone another strategic approach to solve this DS question? Regards!