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by mehrasa » Wed Sep 07, 2011 10:54 pm
if Z is an integer and Z! is divisible by 340, what is the smallest possible value of Z?

Ans: 17

any one can explain this?
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by puneet.goel » Wed Sep 07, 2011 11:04 pm
For Z! to be divisible by 340 ,Z should have all the prime factors of 340.
The factors are : 17*2*2*5.

17 is a prime number and z should include it. so Z! can not be less then 17.
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by navami » Thu Sep 08, 2011 6:05 am
340 can be written as 340 = 4* 5* 17

Now any number Z. Z! to be divisible by 340
so Z! shall contain 340 in it means
Z! = 340 * some other number.
Z! = 4* 5* 17 * Some other number


now imagine Z is 5
Z! = 5 * 4* 3* 2*1
now imagine Z is 4
Z! = 4* 3* 2*1

Z can only be 17 or higher
Z! = 17! will have 17 and also 5 and 4
This time no looking back!!!
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by tpr-becky » Fri Sep 09, 2011 12:56 pm
Factorials are normally an invitation to factor - if you factor out 340 - you get 2(2)(5)(17) this means that z! must be divisible by each of these - since 17 is prime z! must contain a 17 for it to be divisible by 17 - becuase it is a factorial the only way for it to contain a 17 is for z >= 17 thus the smallest z can be is 17.

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