What is the last non - zero digit in (20!)!? (That is, factorial of 20 factorial).

I arrived at an answer of 6 with the following approach:

Let x = 20! = 2432902008176640000

Thus, (20!)! = x! = (x)(x-1)(x-2)…(21)(x)

If we define y = (x-1)(x-2)…(21), then

(20!)! = y*x^2

x^2 has a last non-zero digit of 6 (note the last non-zero digit of 4 in x above).

y, having a very large number of powers of 10, will end in many more zeros than x^2.

Multiplying x^2 by y will, therefore, not affect the last non-zero digit of x^2. Thus, (20!)! has a last non-zero digit of 6.

Posted by NK on 2004-01-23 18:19:54