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!
