Find a three-digit number A such that the digit sum of A is divisible by 12, the digit sum of A + 6 is divisible by 12, and A is greater than 600 and less than 900.

Step-by-step solution

1. 1

   Recall the one fact that governs every problem of this shape. Adding $k$ to a number either leaves the digit sum $k$ larger (no carry) or reduces it by a multiple of $9$ for each carry. Concretely, one units carry changes the digit sum by $k-9$.

2. 2

   Rule out the no-carry case. If $12\mid S$ and no carry occurs, the new sum is $S+6$, which leaves remainder $6$ modulo $12$ — never $0$. So adding $6$ must cause at least one carry, meaning the units digit $b$ satisfies $b\ge 4$.

3. 3

   Work out what one carry does here. With exactly one carry the digit sum changes by $6-9=-3$, giving $S-3$. For $12\mid S$ and $12\mid(S-3)$ to hold at once is impossible, so two carries are needed: the tens digit must also roll over, which requires the tens digit to be $9$. Two carries change the sum by $6-9-9=-12$, and $S-12$ is a multiple of $12$ whenever $S$ is.

4. 4

   Impose the conditions on the digits. Write $A=\overline{hab}$ with $h\in\{6,7,8\}$, $a=9$, and $b\ge 4$. We need $h+9+b\equiv 0\pmod{12}$, so $h+b\equiv 3\pmod{12}$, giving $h+b=15$ (the only value reachable with $h\le 8$, $b\le 9$):

   $(h,b)=(6,9),\ (7,8),\ (8,7).$

5. 5

   List and verify.

   $699\to 6+9+9=24,\quad 705\to 7+0+5=12\ \checkmark$
   $798\to 24,\quad 804\to 12\ \checkmark$
   $897\to 24,\quad 903\to 12\ \checkmark$

   A brute-force scan of $601$ to $899$ returns exactly these three, so any of them is a valid answer.