twosummers Loc: Melbourne Australia or Lincolnshire England

Hello again my friends - our collective UHH brainpower was successful in solving the "everything equals 6" equation puzzle. I also sent it to a couple of clever friends who also struggled with it but got there in the end (one of whom was a Duke maths professor so we should all feel very good about our efforts). However, by way of retaliation the non-Duke professor friend sent me this one which so far has resisted all of my puzzle solving skills - although I am happy to share my logic to approaching the solution. See what you think!

The puzzle is relatively easy to comprehend - take each of the numbers 1 to 9 and make a nine-digit number where - the number in the first position is evenly divisible by 1, the first 2 numbers divisible by 2, the first 3 numbers divisible by 3 and so on.

At first I thought it was easy - start with 1 (that's divisible by 1) then 12 (that's divisible by 2), then 123 (that's divisible by 3) hey this IS easy - now take 1234 and oops (that's not divisible by 4!) and I was going along so well, so back to the start....

At first glance there appears to be an unmanageable number of options to test (9! - there's that pesky factorial again) but wait! - using my ancient and incomplete divisibility rules - the number 5 must be at position 5 as any number ending in 5 or 0 is divisible by 5 and as we don't have to worry about 0 then our first 5 digit piece of the puzzle must end in a 5!! - yippee - I have one digit in the right place.

Then of course the digits in position 2,4,6 and 8 must be even numbers (2,4,6 or 8) as only even numbers are divisible by even numbers. So that leaves the numbers 1,3,5,7 and 9 to occupy the remaining places. I think that reduces the options from a horrendously large number to just a frighteningly large number.

I think I could maybe write a computer program to find the (or a) solution but that would be cheating as would Googling - I see nothing wrong with consulting my photography friends. I am assured that there IS a solution but I'm still wading through it by brute force. I confess to using Excel to check the quotients (is that a word?) and to make matters worse a few times I get to the 8th digit and that fails the divisibility test. What IS it about that pesky number 8?

Good luck my friends - I'm still trying and now I think I'm even retrying old attempts because I'm not keeping track of my failed answers. You have been warned!

TheShoe Loc: Lacey, WA

The first digit and the last can be any digit, so those two cannot be determined until the rest have been established. You already know that any number is divisible by 1 which implies that the first can be any digit. Since all of the digits are u/sed in the result, then no matter what the order, you can cast out nines and determine that the number is divisible by 9.

twosummers Loc: Melbourne Australia or Lincolnshire England

The first and last digit can only be 1,3,7 or 9 (even numbers have to be reserved for positions 2,4,6 and 8. 5 can only be in position 5.

TheShoe Loc: Lacey, WA

Read a little more carefully, please. Do you see the statement that you cannot determine the values of the first and last digits until you have solved all of the others. Each of the 9 digits is divisible by 1 and every 9 digit number made up of all 9 digits, no matter what order, is divisible by 9. You cannot predict either of them based on the description of the first and ninth digits.

steve33 Loc: Yarmouth MA

How about this?
129654783

steve33 Loc: Yarmouth MA

No computer or google search. I used a crayon on the back of an envelope.

steve33 Loc: Yarmouth MA

On further checking, the 8th number doesn't work.

jerryc41 Loc: Catskill Mts of NY

Okay, but was it a magic crayon?

steve33 Loc: Yarmouth MA

Standard crayola

xerxesix Loc: Portland, OR

With a little bit of help from Excel here is a solution: 1 4 7 2 5 8 3 6 9

Daryls Loc: Waco, TX

I got

981654327

Daryl

aflundi Loc: Albuquerque, NM

I think that fails at the 7th digit and 147258369 fails at the 8th . How about 381654729 ?

David in Dallas Loc: Dallas, Texas, USA

That works. How'd you get it?

aflundi Loc: Albuquerque, NM

If you follow the rules already mentioned (digits 1,3,7 and 9 must be from the set [1,3,7,9], digit 5 must be 5, and the rest must be from the set [2,4,6,8]) there's only 576 possible combinations which are a lot easier to sort through and check than 9!=362880 combinations.

xerxesix Loc: Portland, OR

Thanks. I missed that fail on the 8th.