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Question Pack 1 = A positive integer is divisible by 3 if an

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by richachampion » Thu Dec 08, 2016 3:57 pm
A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3. If the six-digit integer is divisible by 3, and n is of the form 1k2,k24, where k represents a digit that occurs twice, how many values could n have?

A. 2
B. 3
C. 4
D. 5
E. 10
Last edited by richachampion on Thu Dec 08, 2016 3:59 pm, edited 1 time in total.
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by richachampion » Thu Dec 08, 2016 3:58 pm
OA: C
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by Matt@VeritasPrep » Thu Dec 08, 2016 7:31 pm
We need (1 + 2 + 2 + 4 + 2k), or (9 + 2k) to be divisible by 3. Since 9 is already divisible by 3, we just need 2k to be divisible by 3, or k to be divisible by 3.

Since k must be a single digit, we could have k = 0, 3, 6, or 9, so there are four options, and we're set!
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by Jay@ManhattanReview » Mon Dec 12, 2016 10:16 pm
Hi Richa,

Just for a change.. What is the answer of the questions had the question been as given below?
A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3. If a six-digit integer n is divisible by 3, and it is of the form k12,k24, where k represents a digit that occurs twice, how many values could n have?

A. Two
B. Three
C. Four
D. Five
E. Ten
-Jay

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