A positive integer has all digits different and none of the digits is 0. If any digit d is dropped from this number, the new number is divisible by d.

Find the greatest number with this property.

Example: 6342 -> 2 divides 634, 3 divides 642, 4 divides 632, 6 divides 342

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    $\begingroup$ Shouldn't there be a "no-computers" tag. Otherwise too easy. $\endgroup$
    – Jens
    Commented Feb 20, 2020 at 12:01
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    $\begingroup$ I really like your questions @ThomasL :) they are pretty smart ones! Thanks. $\endgroup$
    – Oray
    Commented Feb 20, 2020 at 13:48
  • $\begingroup$ @Oray, I'm happy to hear that you like my riddles .. however they are not smart enough for hexomino! $\endgroup$
    – ThomasL
    Commented Feb 20, 2020 at 20:13

2 Answers 2


I think the answer is



The number cannot contain a $5$, since deleting the $5$ means the result cannot end in $5$ or $0$.
If the number contains a $9$, then the sum of its digits is divisible by $9$. $1+2+3+4+6+7+8+9 = 40$ so if we wish the number to contain a $9$, it can have at most $7$ digits. If we want to pursue this path, we would have to exclude $4$ to guarantee divisibility by $9$. This leaves us to construct a $7$-digit number from $1,2,3,6,7,8,9$ with the given property.

The number must end in an even digit to ensure divisibility by each even number and also, the second-last digit must be even to ensure divisibility by the last digit.
If the last two digits are $26$ or $62$ then we cannot get divisibility by $8$. Hence, $8$ must be one of the last two digits.

Now suppose we wish to make the first two digits $97$ to get the maximum number possible. If we require $2$ to be the other one of the last two digits then the fourth and fifth digits must either be $31$ or $63$ to guarantee divisibility by $8$ when $8$ is deleted.
Therefore, the numbers to test in this branch are $9763182, 9763128, 9716382$ and $9716328$. Only the last number passes divisibility by $7$ which is $9716328$ so this is our best answer is this branch.

Hence, we must have that the number begins with the digits $97$ and several of our unexplored branches can be discounted because numbers there will be strictly less than $9700000$.

There remains one branch to explore which is when $6$ is the other of the last two digits (there could be a possible larger solution here). In this case, the fourth and fifth digits must be $21$ or $13$ to ensure divisibility by $8$ when $8$ is deleted.
This leaves four numbers to test: $9732168, 9732186, 9721368, 9721386$.
Again, there is just one candidate which satisfies the divisibility by $7$ rule and that is $9721368$. This is bigger than our previous candidate and hence is the maximum.

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    $\begingroup$ Awesome! Just awesome! Here I was coming up with an algorithm to code it. $\endgroup$
    – Ébe Isaac
    Commented Feb 20, 2020 at 12:47

The answer is

I wrote a Java program to solve it: https://pastebin.com/NSVXnHtr
By the way there is a sequence for this: https://oeis.org/A061362

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    $\begingroup$ Beat me to it +1 $\endgroup$
    – Ébe Isaac
    Commented Feb 20, 2020 at 12:51
  • $\begingroup$ I've simplified my algorithm. $\endgroup$ Commented Feb 20, 2020 at 12:52
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    $\begingroup$ definitely an upvote for the program! $\endgroup$
    – ThomasL
    Commented Feb 20, 2020 at 20:14

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