Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

1x2x3x4x5x6x7……x1990x1991'a answer have a lot of zero at the  end. What is the first number that is not 0 counting from the right?

see: LEGENDRE factorial in prime factors disassemble

1. factor 2

$\begin{array}{rclr} \frac{1991}{2} &=& 995.5 & 995\\ \frac{1991}{2^2} &=& 497.75 & 497\\ \frac{1991}{2^3} &=& 248.875 & 248\\ \frac{1991}{2^4} &=& 124.437 & 124\\ \frac{1991}{2^5} &=& 62.21875 & 62\\ \frac{1991}{2^6} &=& 31.109375 & 31\\ \frac{1991}{2^7} &=& 15.5546875 & 15\\ \frac{1991}{2^8} &=& 7.77734375 & 7\\ \frac{1991}{2^9} &=& 3.888671875 & 3 \\ \frac{1991}{2^{10}} &=& 1.94433593750 & 1\\ \frac{1991}{2^{11}} &=& 0.97216796875 & 0\\ && \text{sum} & = 1983 \end{array}$

2. facor 5

$\begin{array}{rclr} \frac{1991}{5} &=& 398.2 & 398\\ \frac{1991}{5^2} &=& 79.64 & 79\\ \frac{1991}{5^3} &=& 15.928 & 15\\ \frac{1991}{5^4} &=& 3.1856 & 3\\ \frac{1991}{5^5} &=& 0.63712 & 0\\ && \text{sum} & = 495 \end{array}$

...

$1991! = 2^{1983} \times 3^{991} \times 5^{ \color{red}459} \times 7^{329} \times \dots \times 1987^1$

because the exponent of 2 is not less  than the exponent of 5 - 1983 is greater than 459 - we have 459 zeros at the  end of 1991!

You might have some difficulty in understanding "heureka" analysis. Here is a summary of he has calculated: 1991! has 495 trailing zeros. The last 10 or so non-zero digits are:5928553472.

The number has 5706 digits in it.