The Mysterious Code

Throughout this solution I will use the notation (a,b,c) with any number of elements within parentheses to represent all possible permutations of those numbers

Starting with the same method as Satyen:

The sum of the digits $0-9$ is $45$ . The only combination of Prime and >Triangular numbers that works is $17$ and $28$ . One of the first five numerals is $4$ . So the sum of the other four digits is $13$ . The only possible combos for first five and last five digits are

$(0,1,3,4,9)(2,5, 6 ,7, 8)$ ,

$(0, 2, 3, 4, 8) (1 ,5 ,6 ,7 ,9)$ ,

$(0, 1, 4, 5, 7) (2, 3, 6, 8, 9)$ ,

$(1, 2, 3, 4, 7)(0, 5, 6, 8, 9)$ ,

$(0,2, 4, 5, 6)(1, 3, 7, 8, 9)$

The square number can be $1, 4, 9, 16$ however, if it is $9$ then adding the next two digits will not make it big enough to be the next cube $27$ so $9$ is out, the square must be $1, 4, 16$ and the cube will be $8$ if the square is $1, 4$ and $27$ if the square is $16$ .

Take the last case $(0,2, 4, 5, 6)(1, 3, 7, 8, 9)$ :

The only possible square is $1$ so the cube is $8$ and we have $(2,5,6)4013(7,8,9)$ . Clues $3,4$ and $5$ tell us that each number on the left must be $2,3$ or $4$ away from a number on the right, $2$ on the left is not, we have a contradiction and this entire case is thrown out.

Note that if the square is $4$ then we have either $0,4$ or $1,3$ in the middle and as the cube must be $8$ , we use the digits $0,1,3$ and $4$ in the $4$ middle spots. For each case left, $3$ of these digits are part of the left most $5$ and so the square can't be $4$ . For similar reasons, if the square is made of $0$ and $1$ , then the cube must add either $2$ and $5$ or $3$ and $4$ but we just established that in the remaining cases $0,1,3,4$ cannot be in the middle in any order. This means the three possible combinations of middle $4$ digits are $0,1,2,5$ if the square is $1$ and $7,9,3,8$ or $7,9,5,6$ if the square is $16$

In the first case we must have $(0,1,4)3978(2,5,6)$ , the only two that are $4$ apart are $1$ and $5$ so we have $(0,4)3978(2,6)$ but now none are $3$ apart so this case is out

Case $2$ we have $(3,4,8)2015(6,7,9)$ the only two that are $4$ apart are $3$ and $7$ so we have $(4,8)320157(6,9)$ but now none are $3$ apart so this case is also out

Taking case $3$ $(0,1,4)5796(2,3,8)$ the only two that are $4$ apart are $4$ and $8$ so we have $(0,1)457968(2,3)$ however choosing the only digits $3$ apart, $0$ and $3$ for the next digits out, leaves us with digits only $1$ aprat when they must be $2$ .

We now know the number must have the form $(1, 2, 3, 4, 7)(0, 5, 6, 8, 9)$ giving two options:

$(1,2,4)3798(0,5,6) \rightarrow (1,4)237986(0,5)$ because if we choose $1,5$ as the two digits $4$ apart we have only even digits left, none of which could have difference $3$ , however even with the case chosen, no pairs have difference $3$ so this subcase is out.

$(3, 4, 7)2105(6, 8, 9) \rightarrow (3, 7)421058(6, 9) \rightarrow 7342105869$

Adding up the odd digits of $7342105869$ we have $7+4+1+5+6=23$ which is odd so the answer is $\boxed{7342105869}$

Drowning in the casework for this solution, if there is a way to limit it please lemme know!

With these questions, I tend to pick every single possible option that leads to a dead end, before finally getting the correct answer...

Anyway, I have a different approach. We begin with the first two clues. From there, we can shortlist the possible options:

Sum of middle two digits: $1,4,9,16$ (No two distinct digits will give you a sum more than 17)

Sum of middle four digits: $8,27$ (No four distinct digits will give you a sum less than 6 or more than 30)

If sum of squares is $9$ , we cannot get $27$ because the maximum sum of those digits is $26$ . Therefore, we have only three possible cases (time for a case by case analysis!):

(Perfect square, perfect cube) $=(1,8),(4,8),(16,27)$

Note: We try to fit all the numbers first in these cases, the arrangement will be decided later

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Case 1 : $(16,27)$

To get a $16$ , our only possible option is $7+9$ . Our number will then be

$\square\square\square\square79\square\square\square\square$

The two digits beside $7$ and $9$ will need to add up to $11$ . We have these possibilities: $\color{#3D99F6}{5+6},\;\color{#D61F06}{3+8}$

First, we try $\color{#3D99F6}{5}$ and $\color{#3D99F6}{6}$

$\square\square\square\color{#3D99F6}{5}79\color{#3D99F6}{6}\square\square\square$

We are left with $0,1,2,3,4,8$ . To satisfy clues 3, 4 and 5, $8$ must go with $4$ to get a difference of $4$ . $3$ must then go with $0$ to get a difference of $3$ , but that leaves us with $2$ and $1$ . We cannot satisfy clues 3, 4 and 5, so this option is impossible.

We then try $\color{#D61F06}{3}$ and $\color{#D61F06}{8}$ :

$\square\square\square\color{#D61F06}{3}79\color{#D61F06}{8}\square\square\square$

We are left with $0,1,2,4,5,6$ . To satisfy clue 5, we have 3 options:

1. $6$ and $2$ has a difference of $4$ . However, with $0,1,4,5$ remaining, we cannot satisfy clues 3 and 4
2. $5$ and $1$ has a difference of $4$ . However, with $0,2,4,6$ remaining, we cannot satisfy clues 3 and 4
3. $4$ and $0$ has a difference of $4$ . However, with $1,2,5,6$ remaining, we cannot satisfy clues 3 and 4

This implies that Case 1 is impossible. We then move on to the next case

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Case 2 : $(4,8)$

To get a sum of $4$ , we have only two options: $\color{#20A900}{4+0},\;\color{#EC7300}{3+1}$

Note that to get an $8$ , we have to get $4+4$ . This implies that $\color{#20A900}{4,0},\color{#EC7300}{3,1}$ are all in the center, and we have only two possibilities. Let's try out the first one:

$\square\square\square\color{#20A900}{4}\color{#EC7300}{13}\color{#20A900}{0}\square\square\square$

We are left with $2,5,6,7,8,9$ . To satisfy clues 3, 4 and 5, $2$ must go with $5$ or $6$ . If $2$ goes with $5$ , the remaining digits $6,7,8,9$ cannot satisfy clue 5. If $2$ goes with $6$ to satisfy clue 5, $5$ and $8$ will satisfy clue 4, and $7$ and $9$ will satisfy clue 3. We then have

$752\color{#20A900}{4}\color{#EC7300}{13}\color{#20A900}{0}689$

Next, we check to see if it satisfies clues 6 and 7.

Note that adding 5 distinct digits will give a minimum of 10 and a maximum of 35. The triangular numbers between 10 and 35 inclusive are $10,15,21,28$ . The sum of all $10$ digits is $45$ , and these are the possible combinations:

$(10,35),(15,30),(21,24),\color{#27D2E7}{(28,17)}$

Among these possible combinations, only $\color{#27D2E7}{(28,17)}$ satisfies clues 6 and 7.

Now, we return to our number:

$752\color{#20A900}{4}\color{#EC7300}{13}\color{#20A900}{0}689$

The sum of the first $5$ digits is $19$ . We cannot move $4$ as it needs to satisfy clue 9. The other digits in the first half are all the smaller digits, therefore we know that we cannot satisfy clues 6 and 7 with this combination.

Note that the second possibility:

$\square\square\square\color{#EC7300}{1}\color{#20A900}{40}\color{#EC7300}{3}\square\square\square$

will also lead to the same dead end as above. We can then be certain that Case 3 is the only possible option

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Case 3 : $(1,8)$

The two middle numbers must be $0$ and $1$ :

$\square\square\square\square01\square\square\square\square$

To get an $8$ , the two digits beside $0$ and $1$ must add up to $7$ . We have two possibilities: $\color{#69047E}{4+3},\;\color{magenta}{2+5}$

Note that the first possibility

$\square\square\square\color{#69047E}{4}01\color{#69047E}{3}\square\square\square$

will lead to the same dead end we got in Case 2. This implies that the only option we have is

$\square\square\square\color{magenta}{2}01\color{magenta}{5}\square\square\square$

We are now left with $3,4,6,7,8,9$ . To satisfy clue 5, we have 2 options:

1. $3$ and $7$ has a difference of $4$ . However, with $4,6,8,9$ remaining, we cannot satisfy clues 3 and 4
2. $4$ and $8$ has a difference of $4$ . $3$ and $6$ satisfies clue 4, and $7$ and $9$ satisfies clue 3.

Put it in the number:

$734\color{magenta}{2}01\color{magenta}{5}869$

Now, we check and see if it satisfies clues $6$ and $7$ :

From case 2, we know that the first $5$ digits must add up to $\color{#27D2E7}{17}$ , and the last $5$ digits must add up to $\color{#27D2E7}{28}$

The first $5$ digits add up to $16$ , and we have only one way to change this to $17$ : switch the middle numbers

$734\color{magenta}{2}10\color{magenta}{5}869$

$4$ is in the first half, and clue 9 is satisfied. We are left with clue 8:

Sum of digits in odd positions $=7+4+1+5+6=23$ . The sum is an odd number, which satisfies the final clue.

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Punch in the code $\boxed{7342105869}$ , and run towards your freedom!

The sum of the digits $0-9$ is $45$ . The only combination of Prime and Triangular numbers that works is $17$ and $28$ .

One of the first five numerals is $4$ . So the sum of the other four digits is $13$ .

The only possible combos for first five and last five digits is--

$49013, 25678$ ,

$48023, 15679$ ,

$47015, 23689$ ,

$47123, 05689$ ,

$46025, 13789$ .

It is extremely unlikely that $9$ is in the first five digits.

The Cube number can only be $8$ or $27$ and the square number can be $1, 4, 9, 16$ .

Combining the information about difference between digits as provided and sum of odd positions being odd, the only Code that works is $7342105869$