Home > Number theory, Olympiad, Problem Solving > Last non-zero digit of a factorial

Last non-zero digit of a factorial


Find a method to calculate the last non-zero digit of n!, where n \in \mathbb{N}^* and n!=1\cdot 2\cdot ... \cdot n.
Solution: We have the following formula \displaystyle (5q)! ={10}^{q}q!\prod_{i=0}^{q-1}\frac{(5i+1)(5i+2)(5i+3)(5i+4)}{2}, where this is proved by removing from (5q)! terms divisible by 5 : \ 5, 10,...,5q .
Since \displaystyle \frac{(5i+1)(5i+2)(5i+3)(5i+4)}{2} \equiv 2 \text{ mod }10 we can obtain the following reccurence L(n) \equiv 2^q L(q)L(r) \text{ mod }10, where L(n) is the last non-zero digit of n! and n=5q+r by the integer remainder theorem.
This enables us to calculate the last digit of n! very fast, descending exponentially at every step to reach small numbers, for which we can easily calculate that digit.

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  1. September 2, 2011 at 5:12 pm

    Hi,
    Thanks for the method. I can use this as long as r is not equal to 0. What will be the answer when r = 0? Or in other terms, what to do when n is a multiple of 5?
    Regards.

  2. July 27, 2012 at 6:12 am

    why is the residue L(r) ? I suspect there’s a problem when r = 1 since 1 != 6 mod 10, even though 1 * 2 = 6 * 7 , 1 * 2 * 3 = 6 * 7 * 8, 1 * 2 * 3 * 4 = 6 * 7 * 8 * 9 mod 10

    • July 27, 2012 at 11:08 am

      I don’t understand your point. Please detail a bit what you mean. (there is a small error in your comment: 1!=1 \mod 10)

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