Divisible by 99?

First, since the sum of the digits from $1$ to $9$ is divisible by $9$ , any permutation is divisible by $9$ , so we only need to look at divisibility by $11$ . A well known digit test for that is do an alternating sum of the digits, which , if divisible by $11$ , will mean that the number itself is also divisible by $11$ . So, we start with the smallest possible number and check this sum

$1-2+3-4+5-6+7-8+9=5$

which means it's not divisible by $11$ , and we need to make the sum divisible by $11$ . We can change the sum only by transposing some of the digits, and they need to be of a different sign to make a difference. However, the difference in the sum will always be an even number, so it's pointless to do transposes that reduces the sum (can't get from $5$ to $0$ ), so we look for transposes that raises it. We focus on the digits to the far right. We transpose the furthest right digits that will do the job immediately

$1-2+3-4+8-6+7-5+9=11$

and then we rearrange to get the smallest possible number (sort them in each sign group)

$1-2+3-4+7-5+8-6+9=11$

Thus we end up with $123475869$ which is divisible by $99$ .

Relevant wiki: Divisibility Rules (2,3,5,7,11,13,17,19,...)

As the numbers $9$ and $11$ are coprime, we can say that $n$ is divisible by $99$ iff it's divisible both by $9$ and $11$ . As all the numbers consisting digits $1$ through $9$ exactly once are divisible by $9$ (because the sum of the digits is divisible by $9$ ), we are looking for the smallest $n$ consisting each digit exactly once and being divisible by $11$ .

Let's denote the odd position digits of $n$ by $A$ and even position digits by $B$ such as $n = (A_1B_1A_2B_2A_3B_3A_4B_4A_5)\mbox{.}$

The divisibility rule for $11$ tells us that $\sum A - \sum B \equiv 0 \pmod{11}$ . Because the divisibility does not depend on the order of digits within $A$ and within $B$ , the immediate conclusion from the minimality constraint for $n$ is that $A_1<A_2<A_3<A_4<A_5$ and $B_1<B_2<B_3<B_4$ .

Because $\sum A + \sum B = 45$ , we can rewrite the congruent equation as $(45 - \sum B) - \sum B \equiv 0$ . This gives us $2\sum B \equiv 1$ and $\sum B \equiv 6 \pmod{11}$ .

Also we have: $10 = 1+2+3+4 \leqslant \sum B \leqslant 6+7+8+9 = 30$ and this with the congruent equation gives us two solutions: $\sum B = 17$ or $\sum B = 28$ .

Let's look at the solutions where $n$ starts with $1234$ . We have $(B_1, B_2) = (2, 4)$ . If $\sum B = 28$ then $B_3+B_4=22$ and there are no solutions amongst single digits.

If $\sum B = 17$ then we have $B_3+B_4=11$ (and of course $4<B_3<B_4<9$ ). This gives us the only solution $B = (2,4,5,6)$ and the final solution $n=123475869$ .

We have found the only solution starting with the digits $1234$ , thus we can conclude it's the minimal one.

I'm really sorry, I thought this was a Computer Science problem. Though I got how to do it manually, here's the Python code anyway.

import itertools
string = "123456789"
numbers = [int("".join(x)) for x in itertools.permutations(string) if int("".join(x))%99==0]
numbers.sort()
print numbers[0]

Please change the word usage, I did not know that I have to use all the digits from 1 to 9 >_<