Trollathon #1.7 : Gibberish

Using the caesar cipher, we can transform Nodobwsxo dro vkbqocd zycsdsfo sxdoqob x cemr drkd dro pyvvygsxq cdkdowoxd sc dbeo: "dgy dryeckxn kxn pyebdoox pkmdybskv oxnc sx x jobyoc" to Determine the largest positive integer n such that the following statement is true: "two thousand and fourteen factorial ends in n zeroes"

Now, the problem is clear. We just use the legende formula to find out that $2014!$ ends in $\boxed{501}$ zeroes.

If one does not realise it's a Caesar's cipher, but expects an arbitrary mono-alphabetic substitution, it is still quite solvable. First of all, one notices that the word 'dro' occurs several times throughout the text and could expect it to mean 'the'. In that case, 'drkd' must mean 'that'. If one knows that the only two one-letter words in english are 'a' and 'I', one gets confused at this point, since 'x' would have to be 'I'. However, 'pyebdoox' would have to end in 'teei', which doesn't occur in the english language. A next guess could be that 'kxn' means 'and', since 'kxn' both occurs as a word, and as the last three letters of 'dryeckxn'. We now realise that 'x' stands for 'n' and that the word 'sxdoqob' has to be 'integer'. From there, it's an easy exercise to fill in the remaining letters and all that's left to do is count the number of prime factors $5$ in $2014!$ .

use the caesar cipher

Using the Caesar Cipher, we get: Determine the largest positive integer n such that the following statement is true: "two thousand and fourteen factorial ends in n zeroes"

The answer is $\boxed {501}$ .