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

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What is the remainder when a 3 digit number ABC where A , B and C are [#permalink]

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New post 23 Apr 2016, 06:56
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Question Stats:

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What is the remainder when a 3 digit number ABC where A , B and C are its hundreds,tens and units digit respectively is divided by 3
[A] A+B+C = 3K+13 where K is an integer.
[B] A+B+C = 19
[Reveal] Spoiler: OA

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Re: What is the remainder when a 3 digit number ABC where A , B and C are [#permalink]

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New post 23 Apr 2016, 15:53
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Its 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 its is, the remainder will be 0.

Lets test our prethinking :-

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

Lets Examine the facts:-

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

lets 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

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Re: What is the remainder when a 3 digit number ABC where A , B and C are [#permalink]

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New post 03 Oct 2017, 10:43
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Re: What is the remainder when a 3 digit number ABC where A , B and C are   [#permalink] 03 Oct 2017, 10:43
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