# Largest number with no repeated digits that is a multiple of each of its digits

What is the largest positive integer that has no repeated digits and is a multiple of each of its digits?

Obviously, I can write a program to find such numbers by checking every single number, but can I do this mathematically?

• Do you mean none of its digits are repeated? – P Vanchinathan Oct 11 '16 at 12:22
• @PVanchinathan: Yes, I edited the question. My English is not good but is not a downvote for a newcomer too harsh? – user377355 Oct 11 '16 at 12:27
• 9876543210 is divisible by all except 7. Start rearranging the digits from the end and see when the resulting number is divisible by 7. – N.S.JOHN Oct 11 '16 at 13:21
• @N.S.JOHN It's not divisible by $0$ either. – Daniel Fischer Oct 11 '16 at 13:21

Clearly, such an integer cannot contain the digit $0$ (this doesn't depend on the base). The largest such integer has the maximal number of digits among all such numbers. We can't have a nine-digit integer, since if an integer without a $0$ among its digits is a multiple of $5$, its last digit must be $5$, and then it mustn't contain any even digits, which, since repetitions aren't allowed, can give us at most a five-digit integer.
Let's look at integers with more digits. We can't have an eight-digit integer either, since only $1,2,3,4,6,7,8,9$ are left and the sum of these digits is not divisible by $3$. So we can have at most seven digits. Then it must be divisible by $3$, so we must drop one of $1,4$ or $7$. But that leaves $9$ in the set of digits to use, so we must drop $4$ to have a chance of a seven-digit integer without repeated digits which is a multiple of all its digits. Let's try to arrange the digits $1,2,3,6,7,8,9$ so that the resulting integer is divisible by all of them, and as large as possible. Divisibility by $9$ was built in, and if we achieve divisibility by $8$ and by $7$, we are done. Starting with the digits in almost-decreasing order, with the constraint that the last three digits must form a multiple of $8$, we soon find one.
$9876312$ is the largest even integer made up of these digits, and it is divisible by $8$, but unfortunately not by $7$. So let's permute the last digits a bit. $9876132$ isn't divisible by $8$, so we can't have the first four digits be $9876$. Let's try with $9873$ then. That leaves the digits $1,2,6$, to be arranged so that the resulting number is divisible by $8$. The only option for that is $216$, hence we look at $9873216$. This isn't divisible by $7$ either. Next try, start with $9872$. To arrange the digits $1,3,6$ so that the resulting number is divisible by $8$, we have only the option $136$, but $9872136$ isn't divisible by $7$ either. Thus we try to start with $9871$. To get a multiple of $8$, the only candidate is $9871632$, but that is again not divisible by $7$. Thus we move towards numbers starting with $986$, and the largest even such number is $9867312$. Check:
$$9867312 = 9\cdot 1096368 = 8 \cdot 1233414 = 7\cdot 1409616.$$