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Thread: Minimization problem

  1. #1
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    Minimization problem

    Supose $\displaystyle a,b,c$ are positive integers so that $\displaystyle a+b+c=2010$ and $\displaystyle a!*b!*c!=m*10^n$ where $\displaystyle m$ and $\displaystyle n$ are integers and $\displaystyle m$ is not divisible by 10. Which is the minimum value $\displaystyle n$ can have?

    The first thing I did was to find the values a,b,c should have to minimize the multiplication of factorials. Dividing 2010 by 3 = 670, then a=669, b=670 and c=671. Now I stuck, can someone help?
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  2. #2
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    If $\displaystyle p$ is a prime and $\displaystyle t=fcnt(n,p)=\sum_{k=1}^{\infty} \left[\frac{n}{p^k}\right]$, then $\displaystyle n!=p^t M, \quad p \not | M$
    If $\displaystyle p_1 < p_2$, then $\displaystyle fcnt(n,p_1) \geq fcnt(n,p_2)$, thus $\displaystyle n! = M 10^{fcnt(n,5)}$
    For example, $\displaystyle fct(671,5) = 134+26+5+1 = 166$ and so $\displaystyle 671! = a 10^{166}$ where a is not divisible by 10

    I think you can use this to solve the problem
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