This is an oft repeated concept in different examinations but if you can grasp the following algorithm solving problems is a one min task

**Concept**

Lets say D(N) denotes the last non zero digit of factorial, then the algo says

D(N)=4*D[N/5]*D(Unit digit of N)[If tens digit of N is odd]

D(N)=6*D[N/5]*D(Unit digit of N)[If tens digit of N is even]; Where [N/5] is greatest Integer Function

**#Problem 18**

Find the last non zero digit of 26!*33!.

**Solution Scheme and Approach**

D(26)=6*D[26/5]*D(6)=6*D(5)*D(6)=6*2*2=4[D(5) means last non zero digit of 5!=120 which is 2, same for D(6)]

D(33)=4*D[33/5]*D(3)=4*D(6)*D(3)=4*2*6=8

Hence last non aero digit of 26!*33!=4*8=2

Remember

D(1)=1 D(2)=2 D(3)=6 D(4)=4 D(5)=2 D(6)=2 D(7)=4 D(8)=2 D(9)=8

Now solve:

(1)Last Non Zero Digit of 99!*26!.

(2)How many zeros are at the end of 13!+14!+15!+16!+17!+18!

thanks for your post dude

ReplyDelete(1)6 would be the last non zero digit.

ReplyDelete(2)There would be 2 zeros at the end.

Can anybody confirm whether my answers are ryt or not?

10! is 3628800-- hence last non zero digit is 8 right... good, now we will apply ur algorithm

ReplyDeleteD(10)=6*D(10/5)*D(0)=6*D(2)8*0!=6*2*1=12=2

Please let me know if I did something wrong.

Tens digit is odd So D(10)=4*D(2)*D(0)=4*2=8

Delete2nd ka answer 3 zeros ???

ReplyDelete