Assuming N is a fixed abstract number, not merely a notation:

In decimal, the number of trailing zeros is the highest power of 5 (which is rarer than a factor of 2) which divides it.

There is one count for every multiple of 5, which is floor(82/5) = 16. Then one extra for every multiple of 25, so add 3.

There are 19 trailing zeros in the decimal case.

In base 9, pairs of 3's in the prime factorization lead to trailing zeros. That number is floor((floor(82/3) + floor(82/9) + floor(82/27))/2), which is 19 also.

In base 11, floor(82/11) = 7 trailing zeros as no factors of 121 are involved.

The representation 82 itself, before taking the factorial, might be interpreted as a notation in the desired base. If so:

82 in base 9 is 74 in decimal. That would have 17 trailing zeros, the value of floor((floor(74/3) + floor(74/9) + floor(74/27))/2) -- all numbers in the computation being decimal notation.

In base 11, we're looking at what is 90! using decimal notation. It's still not complicated by multiples of 121 being included in the making of the factorial. Floor(90/11) is now 8, as the number of trailing zeros in base 11.