Factorial
Fri, 2009-07-03 17:20
Can anyone tel mw how to solve the below problem
1. 10000! = (100!)K × P, where P and K are integers. What can be the maximum value of K?
a. 102
b. 105
c. 104
d. 103
Mon, 2009-11-02 01:01
#2
thnx 4 d ques ravi
hey incognito...cud u pls xpand on dis ques...actually hw did u use d options in dis...it wud b so lengthy if v use d basic method....
Sun, 2009-12-13 23:07
#3
maximun is maximum
the maximum value of k will be(irrespective of the options given)
(10000!)/(101!)
but from the options it is 105 that is maximum(its the closest value to the actual maximum value)
Thu, 2010-09-16 14:30
#4
ans is 104
10000! will give 2499 zeros by again and again dividing by 5 (see no. system)
100! will give 24 zeros
so
2499/24 will give 104.........
10000! = ( 100!) * ( 101 * 102 * .... 10000 ) which is said to be equivalent to ( 100!) K * P
So, K* P = 101 * 102 * ... 10000.... ( K & P are integers )
maximum value shd be anything.. per the choices it shd be 105 as thats the maximum among choices....
pls let me know if i m wrong.