The Vault Part 1

Fun problem! The process to encode a letter is

1) convert from A-Z to 1-26 (say this gives a value $\alpha$ )

2) compute $\beta=27-\alpha$

3) compute $\gamma=\text{Mod}(\beta-10,26)+1$

The final result gives the encoding.

Given a coded value $\gamma$ , we just reverse the steps:

1) $\beta=\text{Mod}(\gamma+8,26)+1$

2) $\alpha=27-\beta$

3) Convert from 1-26 to A-Z

For example, the code starts with $11$ ; this process gives $11 \to 20 \to 7 \to \text{G}$ . The second character is $9 \to 18 \to 9 \to \text{I}$ , and so on.

The full text is:

GIVEN A NUMBER N AND ITS FACTORIAL K KNOW THAT K CAN BE EXPRESSED AS TWO TO THE POWER OF A MILLION AND EIGHT TIMES A (WITH A MOD TWO EQUALS ONE) AND WHEN WRITTEN IN BASE THREE K HAS FIVE HUNDRED MILLION AND ONE TRAILING ZEROES THE PASSCODE IS THE NUMBER OF TRAILING ZEROES OF K WHEN WRITTEN IN BASE SEVENTEEN

Now to the passcode problem. Unfortunately, the text has an error; the number of trailing zeroes in base $3$ can't be larger than the highest power of $2$ that divides $n!$ . In fact the second "million" in the message should be "thousand".

We have to find a number $n$ such that the highest power of $2$ dividing $n!$ is $1000008$ , the highest power of $3$ dividing $n!$ is $500001$ , and find the the highest power of $17$ dividing $n!$ .

Using the fact that the highest power of $p$ that divides $n!$ is given by $\sum \left \lfloor \frac{n}{p^i} \right \rfloor$

it's not too hard to find that $n=1000017$ , and the passcode is $\boxed{62498}$ .