Question 1091348
<br>The prime factorization of 6 is
{{{6 = 2*3}}}<br>
To be divisible by 6, a number must be divisible by both 2 and 3.<br>
In your problem, the first three digits of a 4-digit number add to 6; to make the number divisible by 3, the last digit could be 0 (the sum of all four digits would still be 6, which is divisible by 3); or it could be 3 (sum of all four equal to 9, also divisible by 3), or it could be 6 (sum now 12, still divisible by 3); or it could be 9 (sum 15, again divisible by 3).<br>
So we get a 4-digit number divisible by 3 if the last digit is 0, 3, 6, or 9.<br>
But the number also has to be divisible by 2, which means the last digit must be even.  So the last digit can't be 3 or 9; and we are left with only two possibilities for the last digit: 0 or 6.