What is the remainder when a 3 digit number ABC where A, B

[AAPL]

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!

[Sionainn@PrincetonReview]

Once you recognize that the sum of the digits need to be divisible by 3, there is no need to plug in for the first statement. The first statement tells us that the sum of the digits will be 13 more than a multiple of 3. Since 13 is not divisible by 3, you know the sum of the digits will not be divisible by 3.

Take care,
Sionann