Sum and product of digits

Let the number $n$ have $k+1$ digits, which are $d_0,d_1,d_2,\dots,d_k$ where $d_k$ is the first digit and $d_0$ is the last.

$\begin{aligned}n&=S(n)+P(n)\\ \therefore d_0+10d_1+10^2d_2+\dots+10^kd_k&=(d_0+d_1+d_2+\dots+d_k)+(d_0d_1d_2\dots d_k)\\ (10-1)d_1+\left(10^2-1\right)d_2+\dots+\left(10^k-1\right)d_k&=(d_0d_1d_2\dots d_k)\end{aligned}$

Taking away some terms from the left hand side (which must all be zero or positive) gives the following inequality:

$\begin{aligned}\left(10^k-1\right)d_k&\leq d_0d_1d_2\dots d_k\\ 10^k-1&\leq d_0d_1d_2\dots d_{k-1}\end{aligned}$

The largest value each of $d_0,d_1,d_2,\dots,d_k$ can take is 9. We can be sure the equality is still satisfied if we let these variables be their maximal value, so:

$\begin{aligned}10^k-1&\leq\underbrace{9\times9\times9\times\dots\times9}_{k\ 9\text{'s}}\\ 10^k-1&\leq9^k\\ 10^k-1&\leq9^{k-1}(10-1)\\ 10^k-1&\leq10^{k-1}(10-1)&\text{since }10^{k-1}\geq9^{k-1}\text{ for }k\geq1\\ 10^k-1&\leq10^{k-1}(10)-10^{k-1}\\ 10^k-1&\leq10^k-10^{k-1}\\ 10^{k-1}&\leq1\\ k-1&\leq0\\ k&\leq1\end{aligned}$ For single digit numbers (i.e. $k=0$ ), $n=S(n)=P(n)$ , and so these are clearly not possible. Therefore the only possibility is where $k=1$ , i.e. 2-digit numbers.

Now that we know the numbers must have 2 digits, we can simplify the above expressions to two variables; $a$ , the first digit, and $b$ , the second digit.

$\begin{aligned}n&=S(n)+P(n)\\ \therefore10a+b&=(a+b)+(ab)\\ 9a&=ab\\ b&=9\end{aligned}$

Therefore we simply require 2-digit numbers that end in a 9, and there are $\boxed{9}$ such numbers: $19,\ 29,\ 39,\ 49,\ 59,\ 69,\ 79,\ 89,\ 99$