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,k42, where k represents a digit that occurs twice, how many values could n have?
Select: A) 1 B) 2 C) 3 D) 4 E) 10
1k2k42   Taking sum of the numericals = 1 + 2 + 4 + 2= 
<gadget id="calculator">1 + 2 + 4 + 2</gadget>
<output>9</output>
9   We require the values of K such that they are also divisible by 3   102042   132342   162642   192942   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 1k2,k42, where k represents a digit that occurs twice, how many values could n have?