Find The 10 Digit Number Classic Puzzle And Solution

I'm thinking of a 10 digit number that uses the each digit 0 to 9 exactly once and where the number formed by the first n digits is divisible by n. What could my number be? The video explains how to find the answer using tests for divisibility.

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[Hloan]

"Did you figure it out ?"

Me : Well... I figured out that the fifth digit was 5 and the last was 0. So, almost !

[janossowski]

Solved it. But after limiting the possibilities to 10, i got lazy, and I thought it would be fun to write a piece of code to do that for me. Obviously I wrote the whole thing, not just checking the 10 numbers. So I spent 2 hours doing that (cuz when u work with numbers and arrays bugs seem to appear 1000 times more often), and got the same answer as I would have in 5 minutes. But damn, the satisfaction

[hadiel-amine]

You should've said: the number formed by the first n digits is divisible by n.

[joeydunn]

Did you figure it out?

Me: Yeah totally. Wait the answer wasn't 1234567890? Oh. well then no...

[panta]

Awesome! I used a slightly different approach, as I used the divisible by 6 and by 8 property before the 4. But I similarly ended up with those 10 numbers.
I would also like to add that the divisible by 9 property we didn't use is automatic if we go for 0 as last digit (I can't remember if you mentioned it), as the sum of the first 9 digits is always 1+2+3+...9=45 regardless of the cyphre order. 45 is divisible by 9, so we know the first 9 digits are divisible by 9.

[plwpahi]

I worked it out to the point where I knew it would come down to figuring out a number divisible by 7, and then watch the rest of the video lol. Great explanation.

[janwillem]

Thnx for helping me to fall asleep, couldn't sleep for like an hour and this works instantly, keep up the great work

[mh]

To me the 'first' digit is the 0, and that really threw me off.

[CodeKujo]

I wasn't sure if the "1st digit" was the 10^9 place or the 10^0 place, so I solved it both ways. They're both interesting problems, but one of them has a LOT more solutions.

[doctyler]

This one kicked my ass, wow I'm going to have to watch this again.

[mmatt]

great puzzle! much harder than usual! keep them coming please presh!

[Sarah-ojbh]

I figured it out. I was feeling ambitious, so I looked up some rules of divisibility by 7 (there are actually several that are more or less practical depending on the number of digits) and did it all using paper and pencil, double-checking the divisibility by 7 with a calculator. Yep, I have no life.

[dokhterpurdyla]

Thanks for uploading this puzzle: I really enjoyed it. I thought I'd cheated when I used brute force from 10 possibilities so I was pretty pleased when I saw you went the same way.

[oqardZ]

I reduced number of combinations to (4!)^2 = 576 and then wrote a simple program to check them all. Nice puzzle.

[MCYodatheGIANT]

This is good. I spent a couple of hours at work one night trying to figure this out a few years ago. I only figured out how to narrow it down to 120 possibilities, and brute forced it from there.

[remivannier]

I loved this puzzle. I must admit I was unlucky enough to forget 8 cases among which was the answer. Thanks a lot !

[pieterluyten]

this is a question that looks like one I had to do on a test to join training weekends for international math olympiads, it was exactly the same but we had to find the one with 6 numbers :).

[pikachudthunderer]

nice... this one was like solving sudoku ;)

[LegendaryFartMaster]

I love this puzzle. I've always had hard time explaining my solution to others tho!
I'm learning from you thanks!!

[pedroloures]

I got this one right. Its incredible how the divisibility rule eleminates a lot of possibilities.