factorials
what is the rightmost non-zero digit of $10000!$ ?
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Re: factorials
I think it is $6$ as $8^{1000} \equiv 6 (mod 10)$
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Nur Muhammad Shafiullah | Mahi
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Nur Muhammad Shafiullah | Mahi
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tarek like math
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Re: factorials
Answer=6.
actually after some factorial last digit of number multiple by last digit of next number. example suppose $x!=A$ and $a$ is last non-zero digit of $A$ and $(x+1)!=B$ and $b$ is last nonzero digit of $B$ and $c$ is last digit of $(x+1)$.
finally i say $b$ is last digit of $a.c$ i try it and it works.
u r right actually when we multiple 5 then it produce zero in pattern and start from other non-zero digit of it's left and then come a re-pattern.
now i try to prove that last digit of n! not periodic. i will give my proof soon u can see it if right.
if anyone find eror otherwise beauty reply it
actually after some factorial last digit of number multiple by last digit of next number. example suppose $x!=A$ and $a$ is last non-zero digit of $A$ and $(x+1)!=B$ and $b$ is last nonzero digit of $B$ and $c$ is last digit of $(x+1)$.
finally i say $b$ is last digit of $a.c$ i try it and it works.
u r right actually when we multiple 5 then it produce zero in pattern and start from other non-zero digit of it's left and then come a re-pattern.
now i try to prove that last digit of n! not periodic. i will give my proof soon u can see it if right.
if anyone find eror otherwise beauty reply it
Last edited by tarek like math on Sat Apr 30, 2011 9:15 pm, edited 1 time in total.
Re: factorials
I think you meant that the sequence of last non-zero digit of $n!$,say $d_n$ is periodic.
But there was a problem proposed by USSR for IMO-1983, which stated that :
Prove that $d_n$ is not periodic.
But there was a problem proposed by USSR for IMO-1983, which stated that :
Prove that $d_n$ is not periodic.
One one thing is neutral in the universe, that is $0$.
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tarek like math
- Posts:56
- Joined:Fri Feb 18, 2011 11:30 pm
Re: factorials
i try to prove last non-zero digit of n! not periodic
suppose 1~n, last non-zero digit of n!is periodic and last non-zero digits are L1,L2,...Ln. suppose A=n!,B=(n+1)!,C=(n+1) and last non-zero digit of A=Ln,B=L1,C=c1. so B=A.C and L1=last digit of (Ln.c1). every time when Ln come as last digit at the end of period next last digit of period will be L1 where L1=Ln.(something like c1). now we prove last digit of C=c1 is not periodic or every time c will not take same value as c1.
C takes value from set of natural numbers means C=1,2,3,...n.(n+1)...etc.
when C=1~10, c=1~9,1; when C=11~20, c=1~9,2; when C=2990~3000,c=1~9,3; when C=3990~4000, c=1~9,4,.
so generally c is not periodic and can take any value of 1~9.
so every c, where c is last digit of (an+1) makes L1 where L1=last digit of Ln.c(here Ln is last digit of an!),not fixed and can take any number among 1~9.
therefore Ln.1,Ln.2,Ln.3,Ln.4,Ln.5,Ln.6,Ln.7,Ln.8,Ln.9 all should produce L1 as last digit. but it is not possible only Ln=5 then c takes maximum 5 numbers not all 1~9.
so 1~n, last non-zero digit of n! not periodic like L1,L2,...Ln.
(i am not 100% sure so reply if anyone find error then i can edit or remove it)
suppose 1~n, last non-zero digit of n!is periodic and last non-zero digits are L1,L2,...Ln. suppose A=n!,B=(n+1)!,C=(n+1) and last non-zero digit of A=Ln,B=L1,C=c1. so B=A.C and L1=last digit of (Ln.c1). every time when Ln come as last digit at the end of period next last digit of period will be L1 where L1=Ln.(something like c1). now we prove last digit of C=c1 is not periodic or every time c will not take same value as c1.
C takes value from set of natural numbers means C=1,2,3,...n.(n+1)...etc.
when C=1~10, c=1~9,1; when C=11~20, c=1~9,2; when C=2990~3000,c=1~9,3; when C=3990~4000, c=1~9,4,.
so generally c is not periodic and can take any value of 1~9.
so every c, where c is last digit of (an+1) makes L1 where L1=last digit of Ln.c(here Ln is last digit of an!),not fixed and can take any number among 1~9.
therefore Ln.1,Ln.2,Ln.3,Ln.4,Ln.5,Ln.6,Ln.7,Ln.8,Ln.9 all should produce L1 as last digit. but it is not possible only Ln=5 then c takes maximum 5 numbers not all 1~9.
so 1~n, last non-zero digit of n! not periodic like L1,L2,...Ln.
(i am not 100% sure so reply if anyone find error then i can edit or remove it)