Find a three-digit number A such that the sum of the digits of A is divisible by 5, the sum of the digits of A + 4 is divisible by 5, and A is greater than 350 and less than 400.

Step-by-step solution

1. 1

   Fix the hundreds digit from the range. Any $A$ with $350<A<400$ starts with $3$, so write $A=\overline{3ab}$ with the tens digit $a\in\{5,6,7,8,9\}$ and the units digit $b\in\{0,\ldots,9\}$. The digit sum is $S=3+a+b$.

2. 2

   Understand what adding 4 does to the digit sum. There are only two cases:

   • If $b\le 5$, no carry happens and the digit sum rises by $4$: $S\to S+4$.
   • If $b\ge 6$, the units digit wraps and the tens digit gains $1$, so the digit sum changes by $4-10+1=-5$: $S\to S-5$.

   This is the crux. If $5\mid S$, then $S+4$ is never a multiple of $5$, but $S-5$ always is.

3. 3

   Conclude that a carry is mandatory. The first case is impossible, so we need $b\ge 6$ — and, to keep the tens digit from carrying too, $a\le 8$ (a tens digit of $9$ would roll over into the hundreds and change the arithmetic again).

4. 4

   Impose the digit-sum condition. With $b\in\{6,7,8,9\}$ and $a\in\{5,6,7,8\}$, require

   $3+a+b\equiv 0\pmod 5.$

   The pairs that work are $(a,b)=(5,7)$ giving $S=15$, $(6,6)$ giving $S=15$, and $(8,9)$ giving $S=20$.

5. 5

   List and check the solutions.

   $A=357:\ 3+5+7=15,\quad 361\to 3+6+1=10\ \checkmark$
   $A=366:\ 3+6+6=15,\quad 370\to 3+7+0=10\ \checkmark$
   $A=389:\ 3+8+9=20,\quad 393\to 3+9+3=15\ \checkmark$

   A brute-force scan of every integer from $351$ to $399$ confirms these are the only three. Any one of them answers the question.