[uphorik_85]

Suppose you want to find the rightmost nonzero digit of 23!

first find the highest powr of 5 in 23 i,e 4
then remove all the factors of 5 from 23! ..let it be (23)!
so..
(23)! =1*2*3*4*6*7*8*9 (units digit is 6)
*11*12*13*14*16*17*18*19 (units digit is 6)
*21*22*23 (units digit is 6)
*1*2*3*4 (units didit is 4) (removing factor of 5 from 5,10,15,20)

units digit of 6^n is alwayse 6
hence units digit of (23)! is 4 i,e (23)!=z4
in odr to get the right most digit of 23! we hav to divide it by 2^4(since 4 is the higst pwr of 5 in 23!)
hence..
let the rightmost non zero digit of 23! is xR
z4=2^4*xR
z4 =16*xR
dis gives R=4..hence rightmost nonzero digit of 23! is 4

[uphorik_85]

Originally Posted by uphorik_85

to find the last non zero digit of any factorial remove all factors of 5 from it..
der ia funda dat..
1.the product of all numbers from 1to 10 not divisible by 5 ends in 6
i,e 1*2*3*4*6*7*8*9 =x6
2.product of all number from 11 to 20 not divisible by 5 ends in 6 i,e
11*12*13*14*16*17*18*19=x6
and so on...
now to get the rightmost non zero digit of a factorial we hav to divide it by the highest per of 5 and the same pwr of 2 ..
lets expand 15!
(15!)'=1*2*3*4*6*7*8*9 (complete)
*11*12*13*14
*1*2*3 (by removing factors of 5 i,e 5,10,15 becomes 1,2,3)
=6*4*6 (unit's digit)
=x4
hence the last digit of (15)'!=x4
so..far we hav divided 15! by 5^3..
now the quotient of division of (15!)' by 5^3 is x4 and is units digtis 4
now we hav to divide x4 by 2^3 tyo get the units digit of 15!
(15!)' %2^3 =R where Ris the rightmost nonzro digit of 15!
x4=8*xR
if R=3 then the units digit is forin the above equation

so the rightmost non zero digit of 15! is 3

i hav made mistake here..
the rightmost nonzero digit of N! N>1 ..ends with either 2,4,6,8..
so in the equation
x4=8*zR
R can be 3 or 8..since R cannot be 3 so its 8.
hence the rightmost digit of 15! is 8

[uphorik_85]

Try dis..
find the last 2 digits of 25!

regards
abhi

[deepakraam]

I did the multiplication of only the last digits from the factorials from 1!-14! but I'm stuck in 14! which I'm getting the last digit as 2.

2*3 = 6
6*4 = 4
4*5 = 2
2*6 = 2....likewise till 14! will give the last digit as 2.but when u multiply with 15,u need to know the last but one digit.So stuck here.

[Whats in a name]

Originally Posted by uphorik_85

Try dis..
find the last 2 digits of 25!

last two non zero digits??? (last two digits = 00)

[hi.shivani]

Originally Posted by uphorik_85

Try dis..
find the last 2 digits of 25!

regards
abhi

I think it asks about last two non zero digits of 25! as the last two digits the other way will be "00".

There are 6 5's in 25!. So to find the last two non zero digits we need to find remainder of 25!/10^8.

[uphorik_85]

Originally Posted by Whats in a name

last two non zero digits??? (last two digits = 00)

yeah its last 2 non zero digits of 25!

thanx for pointing out

[uphorik_85]

And heres another challenger..
which of the following natural nos can be expressed as the sum of squares of six odd integers..

a.1996
b.1997
c.1998
d.1999
e.2000
plz illaborate if its not trial and errror

[manjushaa]

hi.i am unable to solve these P&C problems.can u please try it out and get back to me.
1.in how many ways can the letters of the English alphabet be arranged so that there are 7 letters between A and B?(36.24!)
2.what are the no of natural nos of 2 or more digits in which digits from left to right are in increasing order?(502)
3.How many 6 digit nos contain exactly 4 different digits?(294840)

[aryan007]

Originally Posted by uphorik_85

Suppose you want to find the rightmost nonzero digit of 23!

first find the highest powr of 5 in 23 i,e 4
then remove all the factors of 5 from 23! ..let it be (23)!
so..
(23)! =1*2*3*4*6*7*8*9 (units digit is 6)
*11*12*13*14*16*17*18*19 (units digit is 6)
*21*22*23 (units digit is 6)
*1*2*3*4 (units didit is 4) (removing factor of 5 from 5,10,15,20)

units digit of 6^n is alwayse 6
hence units digit of (23)! is 4 i,e (23)!=z4
in odr to get the right most digit of 23! we hav to divide it by 2^4(since 4 is the higst pwr of 5 in 23!)
hence..
let the rightmost non zero digit of 23! is xR
z4=2^4*xR
z4 =16*xR
dis gives R=4..hence rightmost nonzero digit of 23! is 4

i think answer is 6, here is explanation
23 ! will have 4 multiples of 5, so we'll remove 2 power 4 i.e 16 from 23!
1*2*3*4*5*6*7*8*9---> last digit 6
11*12*13*14*15*17*18*19--->6
21*22*23------------------->6

so last non zero digit is 6