XXXIII Cono Sur Mathematical Olympiad

The largest digit should be equal to the sum of the rest of the digits. We separate in cases according to the value of the largest digit.

Observe that if the largest digit is $1$, then it is the only digit and the number is not happy. In the case that $2$ is the largest digit then either it is the only digit or the number also has a digit $1$; in both cases, the number is not happy. If the largest digit is $3$, we can get $3$ as the sum of smaller digits in exactly one way, namely $1+2=3$. That is, if the other digits are $1$ and $2$ the number is happy. We have $3!=6$ numbers with the digits $1$, $2$ and $3$.

Similarly, since $1+3=4$ there are $6$ happy numbers where $4$ is the largest digit.

If $5$ is the largest digit, we have $1+4=5$ and $2+3=5$ so there are $12$ happy numbers ($6$ with the digits $1$, $4$ and $5$, and $6$ with the digits $2$, $3$ and $5$).

If $6$ is the largest digit, we have $1+5=6$, $2+4=6$ and $1+2+3=6$. We have $6$ happy numbers with the digits $1$, $5$ and $6$, $6$ with the digits $2$, $4$ and $6$, and $4!=24$ with the digits $1$, $2$, $3$ and $6$. That is, we have $6+6+24=36$ happy numbers with its largest digit equal to $6$.

If $7$ is the largest digit, since $1+6=2+5=3+4=1+2+4=7$ we have $6+6+6+24=42$ happy numbers.

If $8$ is the largest digit, since $1+7=2+6=3+5=1+2+5=1+3+4=8$ we have $6+6+6+24+24=66$ happy numbers.

If $9$ is the largest digit, since $1+8=2+7=3+6=4+5=1+2+6=1+3+5=2+3+4=9$ we have $6+6+6+6+24+24+24=96$ happy numbers.

Therefore, in total there are $6+6+12+36+42+66+96=264$ happy numbers.