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1k2,k24 is a number divisible by 3, so 1+2+2+4+2K =multiple of 3:

9+2K= multiple of 3. Since we know that 9 is a multiple of 3, 2K must be also a multiple of 3, so, possible values are:
0, 3,6, and 9.

C) is the answer.
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cf92
Am I the only one who found the question confusing? It asked "how many values could N have" - not K. I understand K can take 4 values, but doesn't any combination of 0, 3, 6, 9 mean a different value for N?

I do not think so, it is clearly stated that the number 1k2, k24, has a digit that occurs twice

best.
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NoHalfMeasures
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

We have a 6 digit integer: 1k2,k24

The 4 given digits sum to 9. This means for this number to be divisible by 3, 9 + 2k must be divisible by three.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

0, 3, 6, and 9 all work. Answer is C.
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An easy question only made difficult in the fact that many people would have ignored the '0'. Nevertheless, practice makes perfect!
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NoHalfMeasures
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 = 9 which is divisible by 3
So 2k can take values which are multiples of 3
So k = 0,3,6,9
Ans: C.4
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Add the digits, the sum comes out to be 9+2k. Plug in values for k from 0 to 9 (k cannot be more than a single digit number), we see the sum comes down to

9+0 - Divisible by 3
9+2
9+4
9+6 - Divisible by 3
9+8
9+10
9+12 - Divisible by 3
9+14
9+16
9+18 - Divisible by 3

Hence the answer is C (4).
NoHalfMeasures
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 n is divisible by 3, and n is of the form 1k2,k24, where k represents a digit that occurs twice, how many values could k have?

A 2
B. 3
C. 4
D. 5
E 10
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The easiest way is to add the digits and sum up. The sum of the digits is 9+2K.
Now, if K=0, sum is 9/3=3 (works)

K=3, 6+9=15/3=5 (works)

K=6, 12+9=21/3=7

K=9, 18+9=27/3=9 so K can have 4 values, 0,3,6,9. Ans is C
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