@Wladmir. If you see carefully,then it is quite obvious that product of n consecutive number will be divisible by n!.
3 must be a divisor of at least one number out of three consecutive number.( because each multiple of 3 lies, 3 number ahead of previous one.) Similarly, we can argue for ith number. that it will be a divisor of at least one number out of i consecutive number. On 6/22/11, Wladimir Tavares <[email protected]> wrote: > @sunny agrawal > In theory, it would be necessary to prove that the product of n consecutive > numbers is divisible by n! > do you agree? > > > Wladimir Araujo Tavares > *Federal University of Ceará > > * > > > > > On Tue, Jun 21, 2011 at 5:01 PM, kartik sachan > <[email protected]>wrote: > >> sir i think the same ans i gave the above.......... >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to [email protected]. >> To unsubscribe from this group, send email to >> [email protected]. >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
