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Question: “Each of the digits O through 9 is used exactly once 10 to create a ten-digit integer. Find the greatest ten-digit number which uses each digit once and is divisible by 8, 9, 10, and 11.”

I have found out so far the divisibility tricks for 8,9,10, and 11.

8: the number the last three digits form must be divisible by 8.

9: the sum of the digits must be divisible by 9.

10: the number must end with a zero.

11: sum the alternating digits. Subtract these two sums. If the result is zero or is divisible by 11, the number is divisible by 11.

I have only found the last 3 digits of the ten-digit so far. This is what I have: _ _ _ _ _ _ _ 2 4 0

Olivier Roche
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Akshaya Gunnam
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1 Answers1

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The condition for 9-divisibility is just trivial because $0 + 1 + 2 + \cdots + 9 = 45$ and $45 \equiv 0$ (mod $9$).

The last 3 digits can also be smaller, e.g. $120, 160, 240$ (the one you suggested), $280, 320, 360, \cdots$.

Edit:

But to find the largest such integer we may just guess from $9876543120$. If you can guess one by one you would find $9876513240$ would be the largest such integer divisible by $8$ to $11$.

Yuki.F
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    There is a larger answer which does end in 240. How the OP could know it ends in 240 without actually knowing the solution is strange. – Jaap Scherphuis Feb 09 '22 at 15:20