Showing posts with label Mathematics. Show all posts
Showing posts with label Mathematics. Show all posts

31 July 2011

IMO problems from 1959 to 2011


IMO problems from 1959 to 2011.
 English language

30 July 2011

Day 1 problem

How many zero are the end of P(10000,13)


Solution
The formula of permutation is P(n, k)=\frac{n!}{(n-k)!}. Then P(10000, 13)=\frac{10000!}{(9987)!}
so we'll find  zero the end of 10000! and 9987!.

there are n_{1}=\left [\frac{10000}{5} \right ]=2000 integers less than 10000 which are multiples of 5,
and n_{2}=\left [\frac{10000}{5^2} \right ]=400 which are multiples of 5^2,
and n_{3}=\left [\frac{10000}{5^3} \right ]=80 which are multiples of 5^3,
and n_{4}=\left [\frac{10000}{5^4} \right ]=16 which are multiples of 5^4,
and n_{5}=\left [\frac{10000}{5^5} \right ]=3 which are multiples of 5^5,

so we get 2000+400+80+16+3=2499 zero are the end of 10000!

By the same way we get 2493 zero are the end of 9987!

So there are 2499-2493=6 are the end of P(10000,13).



Please correct me! by leave your comment . Thanks

29 July 2011

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