Find a three-digit number A such that the digit sum of A is divisible by 4, the digit sum of A + 2 is divisible by 4, and A is greater than 200 and less than 400.

Step-by-step solution

1. 1

   Set up the general form. With $200<A<400$ the hundreds digit $h$ is $2$ or $3$. Write $A=\overline{hab}$ with digit sum $S=h+a+b$.

2. 2

   Work out the effect of adding 2 on the digit sum.

   • No carry ($b\le 7$): $S\to S+2$.
   • Units carry ($b\in\{8,9\}$, $a\le 8$): the units digit drops by $8$ and the tens digit rises by $1$, so $S\to S-7$.
   • Double carry ($b\in\{8,9\}$ and $a=9$): the tens digit also rolls over, so $S\to S-16$.
3. 3

   Eliminate the no-carry case. If $4\mid S$ then $S+2$ leaves remainder $2$, never $0$. So $b\in\{8,9\}$ is forced. In the single-carry case $S\to S-7$, and $S-7$ is a multiple of $4$ only if $S\equiv 3\pmod 4$ — which contradicts $4\mid S$. So the double carry is the only survivor, requiring $a=9$ as well: $S\to S-16$, which stays a multiple of $4$ automatically.

4. 4

   Impose the digit-sum condition with a = 9. We need $h+9+b\equiv 0\pmod 4$ with $b\in\{8,9\}$:

   • $h=2$: $11+b\equiv 0$ needs $b=9$, giving $S=20$;
   • $h=3$: $12+b\equiv 0$ needs $b=8$, giving $S=20$.
5. 5

   List and verify.

   $299\to 2+9+9=20,\quad 301\to 3+0+1=4\ \checkmark$
   $398\to 3+9+8=20,\quad 400\to 4+0+0=4\ \checkmark$

   A brute-force scan of $201$ to $399$ returns exactly these two numbers, so either one answers the question.