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

[richachampion]

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

[richachampion]

OA: C

[Matt@VeritasPrep]

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!

[Jay@ManhattanReview]

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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