# What is the smallest multiple of $999$ with no $9$'s in it? we are considering positive numbers only

What is the smallest multiple of $999$ with no $9$'s in it?

I can't find one by bashing, and I've just been plugging them into a calculator... Are there any smart ways?

Note: Please post a solution that isn't the answer or a bash. I already know the answer, but not how to get it. Thanks!

• why not zero ? no $9$ at all – dato datuashvili Apr 8 '17 at 19:54
• Well, I found that $112 \times 999 = 111888$ with one line of Mathematica code. – Mark McClure Apr 8 '17 at 19:54
• $999*112=111888$ – Nasenhaar Apr 8 '17 at 19:55
• Use that the multiples of 999 are of the form $k\cdot 1000 -k$. – Errol.Y Apr 8 '17 at 19:58
• @Errol.Y You had the crucial idea. Just write an answer, and I will upvote. – Peter Apr 8 '17 at 20:02

Let $X$ be the set of positive integer $n$ such that $999n$ doesn't contain any digit $9$. Let $m$ be the smallest element in $X$ and "$....abc$" be its decimal representation. It is clear $c \ne 0$. Otherwise, $\frac{m}{10} \in X$ contradicts with the choice that $m$ is the smallest element in $X$.
Let "$...def$" be the decimal representation of $999 m = 1000m - m$.
Since $c \ne 0$, it is easy to verify $$(d,e,f) = (9-a,9-b,10-c)$$ Since $999m$ doesn't contain any digit $9$, we have $$\begin{cases} d \ne 9 \implies a \ne 0 \implies a \ge 1\\ e \ne 9 \implies b \ne 0 \implies b \ge 1\\ f \ne 9 \implies c \ne 1 \implies c \ge 2 \end{cases}$$ So the smallest possible candidate for $m$ is $112$. By direct substitution, $999\times 112 = 111888$ doesn't contain any digit $9$. This mean $112$ is the smallest number we seek.