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
Question Pack 1 = A positive integer is divisible by 3 if an
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- richachampion
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Last edited by richachampion on Thu Dec 08, 2016 3:59 pm, edited 1 time in total.
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OA: C
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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!
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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Hi Richa,
Just for a change.. What is the answer of the questions had the question been as given below?
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Just for a change.. What is the answer of the questions had the question been as given below?
-JayA 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
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