Prove that there exists no integer $x > 99$ s.t. $x = \text{product of digits of }x + \text{sum of digits of }x$ I define a number $x$ to be nice if $x = \text{product of digits of }x + \text{sum of digits of }x$. For example,
$$69 = (6 \times 9) + (6 + 9)$$
Motivated by the niceness of $69$, I wish to characterise all nice numbers.
I thought it would be best to slowly increase the number of digits and try to see a pattern.
$2$ Digits
$$x = \overline{ab} = 10a + b = ab + a + b \implies 9a = ab \implies 9 = b \; \text{as} \; a \neq 0$$
$$\therefore x = \{19, 29, 39, ..., 99\}$$
$3$ Digits
$$x = \overline{abc} = 100a + 10b + c = abc + a + b + c \implies 99a + 9b = abc \implies 11a + b = \frac{abc}{9}$$
Testing $0 \le a, b, c \in \mathbb{Z} \le 9$ and $a \neq0$ s.t. $\frac{abc}{9} \in \mathbb{Z}$, I find no solutions exist.

It seems that as we increase the number of digits, no more nice
numbers exist. How do I prove this?

Other interesting questions I thought of:

*

*nice numbers in different base systems?

*considering clusters of digits instead of individual digits?

You don’t have to answer these questions, but if you want to give any insights on them, feel free to do so.
 A: I think it's easy to generalise to two digits in any base.
Let $n = \overline{ab}$ be a two digit number in base $r$, with $a$ and $b$ digits in base $r$ (i.e. $a,b \in \mathbb Z$, $0<a<r$ and $0\leq b<r$).
Then $n = ra+b$, if we want niceness of $n$, we must have $n=ab+(a+b)$, so we get the equation: $ra+b=ab+a+b$ from which we can easily see that $b=r-1$.
The sum of the product and sum of the digits for an $m$-digit number in base $r$ is limited by $m(r-1)+(r-1)^m$, and the $m$-digit numbers in base $r$ are $r^(m-1)$ to $r^m-1$. It's clear that the latter two expression grows faster than the first, so for any base there will be a limit to how many digits nice numbers can have. But for $r=10$ this only allows us to conclude that the number must have fewer than $22$ ($\log_{\frac{10}{9}} 10$) digits.
A: For an n-digit number the product plus the sum of digits is at most 9^n + 9n. If the first two digits are a, b, then x >= (10a+b)10^(n-2) while the sum is less than ab9^(n-2) + a+b+9(n-2).
You can easily show that the sum is less than x when x >= 100.
