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 4k2,k21, where k represents a digit that occurs twice, how many values could n have?	Choose the correct choice from the following answers: A) 2 B) 3 C) 6 D) 5 E) 4
4k2k21   Taking sum of the numericals = 4 + 2 + 2 + 1= 
<gadget id="calculator">4 + 2 + 2 + 1</gadget>
<output>9</output>
9   We require the values of K such that they are also divisible by 3   402021   432321   462621   492921   Answer= 4 = 4
<result>E</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 4k2,k21, where k represents a digit that occurs twice, how many values could n have?