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
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A positive integer is divisible by 3 if and only if the sum
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alanforde800Maximus
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Hi alanforde800Maximux,
We're told that the six-digit integer is divisible by 3, and is of the form 1K2,K24, where K represents a digit that occurs twice. We're asked for the number of different possible values.
Using the 'rule of 3', since 1+2+2+4 = 9 (which is a multiple of 3), the value of 2K (the sum of the other two digits in the number) must ALSO be a multiple of 3. Thus, K could be 0, 3, 6 or 9. That's 4 possible values.
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
We're told that the six-digit integer is divisible by 3, and is of the form 1K2,K24, where K represents a digit that occurs twice. We're asked for the number of different possible values.
Using the 'rule of 3', since 1+2+2+4 = 9 (which is a multiple of 3), the value of 2K (the sum of the other two digits in the number) must ALSO be a multiple of 3. Thus, K could be 0, 3, 6 or 9. That's 4 possible values.
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
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Scott@TargetTestPrep
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We know the following:alanforde800Maximus wrote: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
(1 + 2 + 2 + 4 + 2k)/3 = integer
(9 + 2k)/3 = integer
Thus, k must be a multiple of 3, so it can be 0, 3, 6, or 9.
Answer: C
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