# 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.

• $69$ is a nice number for entirely different reasons that I can't go into on a family-friendly website. :) Jun 28, 2022 at 11:18
• @Deepak Indeed, mathematical inspiration can stem from anywhere!
– user905694
Jun 28, 2022 at 11:19
• This is a finite problem. Should be easy to argue that, for sufficiently large $n$, the product of the digits plus the sum of the digits is $<n$. then it's just a search.
– lulu
Jun 28, 2022 at 11:19
• Small Generalisation: Let $\overline {a9}$ be any m-digit number such that a is an (m—1) digit number. Then $a\times 9+a+9= \overline {a9}$. Eg. $7476487364839= 747648736483\times 9+ 747648736483+9.$ Jun 28, 2022 at 11:22
• Re the modified question: simplistic bounds don't help that much...we get to length $22$ or something like that, too much to search conveniently. But...well, take length $3$. The max of your function is attained at $999$ and we get $756$. But if $n≤756$ then the max is attained at $699$ which yields $510$, and so on. For each length the calculation is easy enough, and it gets easier with greater lengths. Absent a better idea, this should work out quickly enough.
– lulu
Jun 28, 2022 at 12:46

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. $$a,b \in \mathbb Z$$, $$0 and $$0\leq b).

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.

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 $$x \ge 100$$.

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 $$a_1$$.