Math Question - Find the value

12000+10 　

2011-07-24 00:40:54 補充：
(1)
將N! 展開,最後含幾個0,即1,2,3,...,N分解(decomposition)後共有幾個5
so, N!展開最後含0的個數
=[N/5]+[N/5^2]+[N/5^3]+... ( where [x] is the greatest integer function)
In this problem, we should the smalltest integer N, such that
[N/5]+[N/5^2]+... =3000

[N/5]+[N/5^2]+[N/5^3}+.... approaches a geometric series with ratio r= 1/5
so we guess (N/5)/( 1- 1/5 ) = 3000, N=12000

When N=12000, [N/5]+[N/5^2]+[N/5^3]+...= 2400+ 480 + 96+ 19+ 3=2998
3000- 2998= 2,
so the smallest N=1200+10= 1210

(2)
1!=1 is a multiple of 10^0
so, the smallest positive integer N is 1.

2011-07-24 14:13:10 補充：
Sorry! corr.
(1) N=12000+10= 12010

To 意見 004 ,

Your student ?

My student has the following comment:

i) Consider multiple of 5. The question is equivalent to:

[N/5]+[N/25]+[N/125]+... >= 3000

By the trial and error, N = 5*2402 = 12010

ii) 10^0 = 1, all N! is multiple of 1, so smallest positive integer sol. = 1.

Believe?

1) factoring N! into prime factor will give:
N! = P1^(k1) * P2^(k2) * .... * Pm^(km)
where p1 is the first prime (2), p2 is the second prime 3 ... and k1, k2 ... are
their corresponding exponents.
Note that 10^3000 = 2^3000 * 5^3000

Therefore, we know that k1 >= 3000 and k3 >= 3000. Where k1 is the exponent for the prime number P1 which is 2 and k3 is the exponent of P3 which is 5.

It is also obvious k3 > k1 because there are more even number in N! than any number divisible by 5. Every 10 consecutive number will have 2 divisible by 5. So, in order to have the smallest k3 which is 3000, N will have to be
3000*10/2 = 15000. N = 15000

2) 10^0 is 1, the smallest positive integer divisible by 1 is 1.