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 2k1,k24, where k represents a digit that occurs twice, how many values could n have?  Choose the correct answer:	A) 1	B) 2	C) 3	D) 4	E) 10
2k1k24   Taking sum of the numericals = 2 + 1 + 2 + 4= 
<gadget id="calculator">2 + 1 + 2 + 4</gadget>
<output>9</output>
9   We require the values of K such that they are also divisible by 3   201024   231324   261624   291924   Answer= 4 =4
<result>D</result>
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 2k1,k24, where k represents a digit that occurs twice, how many values could n have?