A Maths Puzzle: Find the nine digit number

Find a nine digit numbers, using the numbers 1 to 9, and using each number once without repeats, such that; the first digit is a number divisible by 1. The first two digits is a number divisible by 2; The first three digits is a number divisible by 3 and so on until we get a nine digit number divisible by 9.

This is challenging, don't use a computer for extra points.

Category:

Education

Tags:

• maths
• math
• mathematics
• problem
• puzzle
• number
• theory

[scotland7yard]

cant wait for the solution!

No. 7 is giving me a terrible headache

[singingbanana]

Sounds like you're doing it right then, I found number 7 to be the pain.

[singingbanana]

I wasn't happy with the version I uploaded 12 hours ago. The old version is still there for now.

[agurkas321]

...and why is that so..?

[singingbanana]

There is only one sentence different, but it was important.. Now the audio is worse in this one!

[kdmq]

Thumbs up if you think that all the thumbs ups on the comment of the correct solution places it at the top of the page and ruins it for everyone.

Also thumbs up to place this comment at the top of the page so that this puzzle isn't ruined for everyone!

[jonoiskool]

YEAH!!! 381654729

zomg only took me 2 hours n i didn't cheat TYVM! time to listen to some music!

[chinogh]

1 digits = 1, 2 digits = 18, 3 digits = 189, 4 digits = 1896, 5 digits = 18965, 6 digits = 189654, 7 digits = 1896545, 8 digits = 18965456, 9 digits = 189654561, 10 digits = 1896545610, 11 digits = 18965456103, 12 digits = 189654561036, 13 digits = 1896545610362, 14 digits = 18965456103622, 15 digits = 189654561036225, 16 digits = 1896545610362256,

[BeastOfTraal]

the 5th digit has to be 5 even digits have to be even odd digits are odd. the first two digits have 4 possible choices the next 2 digits have 3 possible choices you know the 5th digit so one choice

the next 2 digits have 2 possible choices and the last two digits only have one choice left so

4 * 4 * 3 * 3 *1 * 2 * 2 *1 * 1 = 576 viable numbers to try

[PoketoMtg]

@PoketoMtg I'ld say it took me 10 minutes as I know most of the rules of multiplication

[PoketoMtg]

381654729 while watching another video by you. First things first the odds go in odds, even go in evens. You can assume is 5 and 7 go into the 5th and 7th spots respectively (multiples of 5 end with 5, and doubleing rule when dividing by 7). Next you find that 2 and 6 fill the 4th and 8ths spots as odd(2/6) are the only ways to make multiples of 4. Putting 8 in the second spot you need a 1 in the 1st or 3rd, to make a multiple of 3 and the last spot is other (4 cannot go cause 1+4=2mod3).

[alancheira]

@HaslamCorp @HaslamCorp I got the same result. I narrowrd down using the 5 first, also, and then the divisibility per 2, then per 8, and went backwards. Luckily the 2 try was divisible by seven at the end, that was my fear. :)

[AudioJustG]

I couldn't come up with a way to solve it myself. So I wrote a program with Delphi to check every single number manually. Sorry for cheating, I'm bad at math, but I wanted to know the answer without looking at comments or anything :(

[jofer123062]

381654729 in less than 30 mins. the trick is simply to know your divisibility rules for each number.

[PukeOfThePukes]

THERE IS A CURSE!!! GIVE ME A CALCULATOR!!! NOWWWWWW