"We're looking for a number with 7 unique digits. The sum of the first seven digits is 40. The sum of the seven middle digits is 30, the product is 0. The sum of the five middle digits is 25, the product is 448. The sum of the last 7 digits is 32, the product is 6400. What are the possible numbers we're looking for?"

Great, so let's solve this in the most inefficient way possible.

000 000 0 - fail
000 000 1 - fail

hm, this will take a while. Maybe someone will be able to do it faster then me :P

What are the possible numbers we're looking for?

Do you mean the unique digits or the number, or numbers made up of those unique digits?

It would matter how long that number made up of the unique digits was.

I am satisfied that I know what the unique digits are.

4

We're looking for a number with 7 unique digits. The sum of the first seven digits is 40. The sum of the seven middle digits is 30, the product is 0. The sum of the five middle digits is 25, the product is 448. The sum of the last 7 digits is 32, the product is 6400. What are the possible numbers we're looking for?

8790178185541

Digits: 0 1 4 5 7 8 9 (7 uniques)
First 7: 8 7 9 0 1 7 8 (sum: 40)
Middle 7: 0 1 7 8 1 8 5 (sum: 30; product: 0)
Middle 5: 1 7 8 1 8 (sum: 25; product: 448)
Final 7: 8 1 8 5 5 4 1 (sum: 32; product: 6400)

Just one of the possibilities (I'm going to leave most of the work undone and turn this in for partial credit so I don't get nerd-sniped any further). To generate the others, you can rearrange some digits (like 879) in here.

Middle 5 have to be 1, 1, 7, 8, 8 (other factorizations of 64 do not add up to 25). Then you need a 0 on one side and a 5 on the other, since the sum of the middle 7 is 30 (25+5) but the product is 0 (so one of the digits has to be 0). Since 6400 is a multiple of just 2's and 5's, the last 7 digits need to be 4, 8, 8, 5, 5, and 2 1's. So the 5 has to go on the right of the middle 5. The first 3 are relatively easier since you just need to throw in 3 numbers that equal 24 once you've moved the appropriate digits in the middle 5 out o the first 7.

@Tabby: Are you sure of that? I think the restrictions placed here might be sufficient to reduce this to a finite subset of integers.

Edited 12/15/2017 04:31:54

Wouldn't it still be countably infinite? If you can shove in random positive integers between the first 7, middle 7, and final 7, then a bijection should exist between solutions generated using that approach and the set of rational numbers.

I guess the ability to throw in arbitrarily long strings of zeros before either of those buffer integers would make it uncountably infinite.

(answered)

We can still describe all members of this set just by extrapolating from Sarah's initial constraints.

Edited 12/15/2017 05:28:06

First -> 40
Middle -> 30, 0 ; 25, 448
Last -> 32, 6400

First -> 9,8,7,6,5,4,1 = 40
-> 9,8,7,6,5,3,2 = 40

Second -> 0,X,X,X,X,X,y
-> sum of seven - sum of five = 30 - 25 so 5.
-> 0,X,X,X,X,X,5
-> Factors(448) => 2*2*2*2*2*2*7
-> 0,7,X,X,X,X,5
-> 0,7,2,4,8,1,5

Last -> 32;6400
-> Factors(6400) => (5*2)*(5*2)*(2*4*8)
-> 4 short ???
-> 5,5,4,8,8,1,1

non-unique digits (?)

If we allow the first, middle, and last to overlap I believe the shortest solution would be 17 digits long.

For all solutions the number of sets of unique digits that would satisfy the conditions is 4, just as Bane hinted at.

8,7,5,4,1,0 & a seventh digit which could be 9,6,3, or 2.

{9,8,7,5,4,1,0}
{8,7,5,4,3,1,0}
{8,7,6,5,4,1,0}
{8,7,5,4,2,1,0}

So it seems to me.

Edited because at 2:30 AM I should have been sleeping - heck!

Edited 12/15/2017 17:48:18

Aaah, it is a single number made of a set of 7 digits. Each digit can be repeated. And all the 3 set of 7 digits can overlap as they all have the same 7 digits... Right ?

Well, I think the last set of digits is pretty much one solution if want to have product 6400 and 32 sum.

Edited 12/15/2017 14:23:46

I'm feeling too smart for this.

Farah dont cheat. For everyone who has never seen this. This is a problem from the aivd (Dutch secret service) christmas puzzle.
Farah, try to connect it with what remains
from ABDEJRSTU FKNPS DGHIKLRT BDGHIJNPRST BGINPRSTU ABERU DEKLNRSVY

I solved it days ago derfellios ;) Also, this isn't the exact same puzzle as the AIVD gave us. They gave another constraint and that puzzle got me wondering about general cases of such puzzles with less constraints. For anyone wanting to solve the puzzle with the last constraint: the product of the first seven digits has to be zero. But that one's boring :)

Edited 12/16/2017 10:05:22

I have not tried to solve that one. But good luck with the other questions!

Edited 12/16/2017 10:55:51

Thanks! We have about 22 points so far, but man oh man this year's edition is hard

This edition is really difficult and long and less points for each question. We have about 10 points so far but I dont think I can reach the 70 we had last year

70, that's pretty impressive! I think we had about 60 last year. Good luck keeping that record up :)

Edited 12/16/2017 10:43:18

I solved it days ago derfellios ;) Also, this isn't the exact same puzzle as the AIVD gave us.

You work in the AIVD?

Man, Dutch intelligence must be seriously underfunded if they have to crowdsource solutions from Warzone.