296. The difference between a number and the product of its digits is equal to the sum of the digits of this number. Find this number.

## Chapter $I X$.

## VARIOUS PROBLEMS
296. Let the desired number contain $n$ digits $a, b, c$, $\ldots, k, l$. According to the condition $a \cdot 10^{n-1}+b \cdot 10^{n-2}+\ldots+$ $+k \cdot 10+l-a b c d \ldots k l=a+b+c+\ldots+k+l, \quad$ from which $9 k=a\left(b c \ldots k l+1-10^{n-1}\right)+b\left(1-10^{n-2}\right)+c\left(1-10^{n-3}\right)+$ $+\ldots(1)$.

In the general case, this equality is impossible. Indeed,

$$
b c \ldots k l-\left(10^{n-1}-1\right)<10^{n-1}-\left(10^{n-1}-1\right)
$$

$b c \ldots k l-\left(10^{n-1}-1\right)<1$; since $b c \ldots k l-\left(10^{n-1}-1\right)-$ is an integer, not equal to zero, it is negative; thus, the right side of equality (1) is negative, while the left side is non-negative.

The only possible case is $n=2$ (by the problem's meaning $n \neq 1$), then $10 a+b-a b=a+b$, from which $b=9 ; a-$ is any digit (not zero). The problem is satisfied by two-digit numbers ending in 9.