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Math Help - Minimization problem

  1. #1
    Junior Member
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    Minimization problem

    Supose a,b,c are positive integers so that a+b+c=2010 and a!*b!*c!=m*10^n where m and n are integers and m is not divisible by 10. Which is the minimum value 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 p is a prime and t=fcnt(n,p)=\sum_{k=1}^{\infty} \left[\frac{n}{p^k}\right], then n!=p^t M, \quad p \not | M
    If p_1 < p_2, then fcnt(n,p_1) \geq fcnt(n,p_2), thus n! = M 10^{fcnt(n,5)}
    For example, fct(671,5) = 134+26+5+1 = 166 and so 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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