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
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.
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....
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)
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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.
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....
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)
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