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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!
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!
