Find a three-digit number A such that the digit sum of A is divisible by 6, the digit sum of A + 3 is divisible by 6, and A is greater than 350 and less than 400.

Step-by-step solution

1. 1

   Fix the form of the number. From $350<A<400$ the hundreds digit is $3$, so $A=\overline{3ab}$ with $a\in\{5,6,7,8,9\}$ and digit sum $S=3+a+b$.

2. 2

   Work out how adding 3 changes the digit sum. Two cases only:

   • No carry ($b\le 6$): the units digit gains $3$, so $S\to S+3$.
   • Carry ($b\ge 7$): the units digit drops by $7$ and the tens digit gains $1$, so $S\to S+3-10+1=S-6$.
3. 3

   Rule out the no-carry case. If $6\mid S$ then $S+3$ leaves remainder $3$ on division by $6$, never $0$. So a carry is compulsory and

   $b\in\{7,8,9\}.$

   In the carry case $S\to S-6$, which is automatically a multiple of $6$ whenever $S$ is — so the second condition becomes free.

4. 4

   Impose the digit-sum condition. Require $3+a+b\equiv 0\pmod 6$ with $a\in\{5,\ldots,9\}$ and $b\in\{7,8,9\}$. The pairs that work are $(a,b)=(6,9),(7,8),(8,7)$, each giving $S=18$.

5. 5

   List and verify the answers.

   $369\to 3+6+9=18,\quad 372\to 3+7+2=12\ \checkmark$
   $378\to 18,\quad 381\to 12\ \checkmark$
   $387\to 18,\quad 390\to 12\ \checkmark$

   A scan of every integer from $351$ to $399$ confirms there are no others.