The product generated by 1 x 2 x 3 x ..... x 10 (base 10), when expressed in base 12, has k zeros. The value of k is:
I followed a similar example '' Determine the number of zeros in which the product generated by 1 x 2 x 3 x ... x 555 ends when it is written in base 6 '' knowing that from this example 2 (6) x 3 (6) = 10 (6) I divided 555 (6) by 3 (6) and did that base division, I added the quotients (in this case were 155 (6) + 35 (6) + 11 (6) + 2 ( 6)) and yielding 251 (6) = 103 zeros. It can be solved following the same logic (prime number theory).