Prove that there exists no integer $x > 99$ s.t. $x = \text{product of digits of }x + \text{sum of digits of }x$

Question

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.

Answer 1

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 expressions grow 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.

Answer 2

For an $n$-digit number the product plus the sum of digits is at most $9^n + 9n$. If the first two digits are $a$ and $b$, then $x \ge (10a+b)\cdot10^{n-2}$ while the sum is less than $ab\cdot9^{n-2} + a+b+9(n-2)$.

You can easily show that the sum is less than $x$ when $x \ge 100$.

Answer 3

You need $\sum_{r=0}^{n-1}10^ra_r = a_{n-1}\prod_{r=0}^{n-2}a_r+\sum_{r=0}^{n-1}a_r$, so $$a_{n-1}\Big(10^{n-1}-\prod_{r=0}^{n-2}a_r-1\Big)+\sum_{r=1}^{n-2}(10^r-1)a_r = 0.$$ But every term on the left hand side is positive for $n>2$ since $10^{n-1}-1$ is larger than the product of $n-1$ digits all less than $10$ (except for $n=2$). This argument works in every base: just replace $10$ by $b$ to see that $n$ cannot be larger than 2, and if $n=2$, $a_0=b-1$, with any $a_1$.