Question 1091348: From left to right, the first three digits of a 4-digit number add up to 6. Which digit could be in the ones place if the 4-digit number is divisible by 6?
Found 3 solutions by ankor@dixie-net.com, MathTherapy, greenestamps:
Answer by ankor@dixie-net.com(22740)   (Show Source):
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
From left to right, the first three digits of a 4-digit number add up to 6. Which digit could be in the ones place if the 4-digit number is divisible by 6?
Easiest one, yet: 
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The prime factorization of 6 is
To be divisible by 6, a number must be divisible by both 2 and 3.
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).
So we get a 4-digit number divisible by 3 if the last digit is 0, 3, 6, or 9.
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.
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