Question 248626
Let's call the first and last digit "x" and the middle two digits "y". So our 4-digit number is in the form: xyyx. In order for this to be divisible by three, the digits must add up to a number divisible by three. So:
x+y+y+x = 2x + 2y = 2(x+y)
must be divisible by three. I wrote it in factored form because it gives us a clue to the solution. Since 2 is not divisible by 3, it must be the (x+y) which is divisible by three. And what are the two small numbers that add up to something divisible by 3? Answer: (1 and 2) or (3 and 0)<br>
So our possible 4-digit numbers are:
1221
2112
3003
(Note: x cannot be zero or else the number would not actually be a 4-digit number). As we can see, the smallest of these is the first one: 1221.