Safe

[cinelli]

Unfortunately I have lost the combination to my safe. What I can remember, however, is that is consists of 16 digits and that they are arranged as follows:

there are two 8s separated by 8 digits,
two 7s separated by 7 digits,
two 6s separated by 6 digits,
two 5s separated by 5 digits,
two 4s separated by 4 digits,
two 3s separated by 3 digits,
two 2s separated by 2 digits,
two 1s separated by one digit,
and that it ends 528.

What is my combination?

Cinelli

[AleisterCrowley]

Can you clarify x digit separation? 3***3 or 3**3?

[ReformedCharacter]

Unfortunately I have lost the combination to my safe. What I can remember, however, is that is consists of 16 digits and that they are arranged as follows:

there are two 8s separated by 8 digits,
two 7s separated by 7 digits,
two 6s separated by 6 digits,
two 5s separated by 5 digits,
two 4s separated by 4 digits,
two 3s separated by 3 digits,
two 2s separated by 2 digits,
two 1s separated by one digit,
and that it ends 528.

What is my combination?

Cinelli

Spoiler alert ...
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

4 7 1 6 1 4 8 5 3 7 6 2 3 5 2 8

RC

[Gengulphus]

Unfortunately I have lost the combination to my safe. What I can remember, however, is that is consists of 16 digits and that they are arranged as follows:

there are two 8s separated by 8 digits,
two 7s separated by 7 digits,
two 6s separated by 6 digits,
two 5s separated by 5 digits,
two 4s separated by 4 digits,
two 3s separated by 3 digits,
two 2s separated by 2 digits,
two 1s separated by one digit,
and that it ends 528.

What is my combination?

Spoiler...

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Given AleisterCrowley's question, let's try it both ways:


n-digit separation means there are (n-1) digits strictly between the two ns

We can straightforwardly fill in where the other 5, 2 and 8 must be - the solution must be of the form:

.......85...2528

This leaves:

* 8 possibilities for where the two 1s can be next to each other
* 5 possibilities for where the two 3s can have 2 digits between them
* 5 possibilities for where the two 4s can have 3 digits between them
* 4 possibilities for where the two 6s can have 5 digits between them
* 3 possibilities for where the two 7s can have 6 digits between them

To minimise the number of possibilities we need to explore, split into cases for where the 7s are placed:

..7....857..2528 (3, 4 or 6 in 3 ways)
...7...85.7.2528 (6 in 2 ways)
....7..85..72528 (6 in 2 ways)

In each case, I've appended a note about which digit has the fewest possibilities for where it is next placed and how many possibilities there are. By choosing to list the cases for where the 6 is placed next, we get seven cases and still have the 1s, 3s and 4s to place in each case:

6.7...6857..2528
..7.6..8576.2528 (*)
..7..6.857.62528
6..7..685.7.2528 (*)
...7.6.85.762528
6...7.685..72528
...67..856.72528

Now look at the rightmost unfilled space in each of those cases. In the two asterisked cases, none of 1, 3 and 4 fit, so we can eliminate those cases. In the remaining five cases, only one of 1, 3 or 4 fits, so we can fill the one that fits in there, list the numbers still to be fitted in for each case and whether / in how many ways that can be done:

6.7...6857112528 (3s and 4s still to be filled in - doesn't work)
..7..64857462528 (1s and 3s still to be fitted in - doesn't work)
...7.63853762528 (1s and 4s still to be fitted in - works in one way)
6...7.6851172528 (3s and 4s still to be fitted in - doesn't work) 
...67.4856472528 (1s and 3s still to be fitted in - works in one way)

So that gives two possible solutions if n-digit separation is to be interpreted in this way:

4117463853762528
1136734856472528

n-digit separation means there are n digits strictly between the two ns

Again, we can straightforwardly fill in where the other 5, 2 and 8 must be - the solution must be of the form:

......85...2.528

This leaves:

* 7 possibilities for where the two 1s can have 1 digit between them
* 5 possibilities for where the two 3s can have 3 digits between them
* 4 possibilities for where the two 4s can have 4 digits between them
* 4 possibilities for where the two 6s can have 6 digits between them
* 4 possibilities for where the two 7s can have 7 digits between them

That leaves a somewhat arbitrary choice of whether to split into cases for where the 4s, 6s or 7s are placed - I'll go for the 7s:

7.....857..2.528 (3 or 4 in 2 ways)
.7....85.7.2.528 (3, 4 or 6 in 3 ways)
..7...85..72.528 (1 in 2 ways)
....7.85...27528 (3 in 2 ways)

In each case, I've appended a note about which digit has the fewest possibilities for where it is next placed and how many possibilities there are. For the third case, place the 1s next, resulting in 2 subcases:

.171..85..72.528
..71.185..72.528

In each of those subcases, there is then only one way to place the 6s:

.171.685..726528
.671.1856.72.528

and it is then easy to see that there is no way to place the 3s and the 4s in either of them.

Returning to the other three cases, we place the 3s next for each of them, resulting in seven cases, in each of which we have the 1s, 4s and 6s still to place:

73...3857..2.528
7....38573.2.528 (*)
37..3.85.7.2.528
.7..3.8537.2.528
.7....8537.23528
.3..7385...27528
....7385.3.27528

In the asterisked case, there is no way to place the 4s, and in each other case, there are only one or two ways to place then - enumerating the cases that leaves, we have:

73..438574.2.528
37.43.8547.2.528
37..3485.742.528
47..348537.2.528
.7..34853742.528
47...48537.23528
.7...48537423528
.3.473854..27528
...4738543.27528

It is then easy to go through those nine remaining cases and see that it's only possible to place the 1s and 6s correctly in one of them, resulting in the solution:

4716148537623528

Since the solution to such problems is supposed to be unique, I infer that n-digit separation is supposed to mean n digits strictly between the two ns, and that the combination is 4716148537623528 - but if that's an inference too far, it won't take me long to try 4117463853762528 and 1136734856472528 as well!

Gengulphus

[cinelli]

4716148537623528 was the combination I was looking for. ReformedCharacter gave the first answer. I wonder what his method was. Gengulphus produced a comprehensive solution so congratulations to both.

Cinelli

[ReformedCharacter]

4716148537623528 was the combination I was looking for. ReformedCharacter gave the first answer. I wonder what his method was. Gengulphus produced a comprehensive solution so congratulations to both.

Cinelli

Thank you very much for the puzzle.

Lacking the brain cells and clarity of thought of Gengulphus, I had to choose an easier method. I used a spreadsheet and using 16 columns put 528 in the appropriate columns. I then made groups on separate rows with 2 x 7 separated by 7 cells, 2 x 6 separated by 6 cells etc. It was just a matter of sliding the groups around until they 'fitted', that provided the answer quite quickly.

RC

[AleisterCrowley]

That's what I was doing....!