wat is the last non zero digit of 25! and 35!
viveknitw Sayswat is the last non zero digit of 25! and 35!
last non-zero digit of 20! = (4^2) * (4!) = 4...
25! = 20! * 21*22*23*24*25 = 4
last non-zero digit of 30! = (4^3) * (6!) = 8..
35! = 30! * 31*32*33*34*35 = 2
kewl
u r correct man.
answer is none of these
(2^31) - 1 by itself is a prime number .. i dont think there can be any prime factors
correct me if am wrong
Well. All I can say is 2^31 -1 is not a prime number and answer is indeed among the options.
The prime factor of (2^31) - 1 lying between 200 and 300 is :
a)227 b)223 c)251 d)271
answer is none of these
(2^31) - 1 by itself is a prime number .. i dont think there can be any prime factors
correct me if am wrong
well i am also getting the same answer as NAGA....none of these..i think 2^31 - 1 is a prime number...any comments...definitely not getting from these options....
jain_ashu SaysWell. All I can say is 2^31 -1 is not a prime number and answer is indeed among the options.
2^odd - 1 is a prime number .. thats what i say .. let me know the answer if u have ..
ps : i have now even checked in normal calculator if ma logic is wrong .. its surely a prime .. pls enlighten me that its not prime number
last non-zero digit of 20! = (4^2) * (4!) = 4...
25! = 20! * 21*22*23*24*25 = 4
last non-zero digit of 30! = (4^3) * (6!) = 8..
35! = 30! * 31*32*33*34*35 = 2
hi,
can you please explain the funda here?
thanks!
divishth Sayswell i am also getting the same answer as NAGA....none of these..i think 2^31 - 1 is a prime number...any comments...definitely not getting from these options....
2^odd - 1 is a prime number .. thats what i say .. let me know the answer if u have ..
ps : i have now even checked in normal calculator if ma logic is wrong .. its surely a prime .. pls enlighten me that its not prime number
Ok. Well I wanted to share this bit of theorem that I found out, with you people, but before that I wanted to know if we have any other approach.
Theorem says,If P is an odd prime number then each prime
divisor of 2^P - 1 is of the form (2*P*k)+1.
Therefore, each prime divisor of 2^37 - 1 is of the form 74k+1.
To get a value between 200 and 300 put k = 3, 74*3 + 1 = 223 is the number.
P.S : Sorry for teasing but I think its worth it.
It seems to be working only if P>19
hi,
can you please explain the funda here?
thanks!
to get the last non zero digit of a number N d formula to be applied is (4^n)*(2n)!
(for N=10,20,30.....) where n=N/10
no for 25!
we take 20 nd go ahead to 25
so for 20 it is (4^2)*(2*2)!=4
now 25!=25*24*23*22*21*20!=4
hope it is clear now.........
Ok. Well I wanted to share this bit of theorem that I found out, with you people, but before that I wanted to know if we have any other approach.
Theorem says,If P is an odd prime number then each prime
divisor of 2^P - 1 is of the form (2*P*k)+1.
Therefore, each prime divisor of 2^37 - 1 is of the form 74k+1.
To get a value between 200 and 300 put k = 3, 74*3 + 1 = 223 is the number.
P.S : Sorry for teasing but I think its worth it.
It seems to be working only if P>19
thats useful .. but then u could have given 2^37 - 1 instead of 2^31 -1
naga25french Saysthats useful .. but then u could have given 2^37 - 1 instead of 2^31 -1
yep I made mistake. Serious typo error
hi,
can you please explain the funda here?
thanks!
well
lets take 100!
Z(100) = L(4^10)*Z(20)\\
every 10! has 4 as last digit so last digit of 4 raised to 10 * number o fives present i. e 20
the process goes on.......................until v simplify
Z(20) = L(4^2)*Z(4)
Z(4) = 4 and L(4^2) = 6
so Z(20)= L(4*6) = 4
L(4^10) = 6
so Z(100) = L(4^10)*Z(20) = 6*4 = 4......
hope i m c
learfeel free to ask
jain_ashu Saysyep I made mistake. Serious typo error
we cud atleast hv done smthing had the question been 2^37 -1
going by options only though:-
2^37 - 1 = 32x^4 - 1, x = 2^8
223 = 2^8 - 33
32 * 33^4 mod 223 = 32 *26^2 mod 32 = 32 *7 mod 223 = 1
final remainder = 0
so 223 is a factor.
Thanks for the theorem though.
we cud atleast hv done smthing had the question been 2^37 -1
going by options only though:-
2^37 - 1 = 32x^4 - 1, x = 2^8
223 = 2^8 - 33
32 * 33^4 mod 223 = 32 *26^2 mod 32 = 32 *7 mod 223 = 1
final remainder = 0
so 223 is a factor.
Thanks for the theorem though.
which therom ?
:shocked:
Ok. Well I wanted to share this bit of theorem that I found out, with you people, but before that I wanted to know if we have any other approach.
Theorem says,If P is an odd prime number then each prime
divisor of 2^P - 1 is of the form (2*P*k)+1.
Therefore, each prime divisor of 2^37 - 1 is of the form 74k+1.
To get a value between 200 and 300 put k = 3, 74*3 + 1 = 223 is the number.
P.S : Sorry for teasing but I think its worth it.
It seems to be working only if P>19
which therom ?
:shocked:
Refer above. Cheers!!!
Puys,I m new to Fermat Theorem..But plz help..
Fermat theorem : for any prime number p , (p-1)times,same digit is repeated , then that number formed is exactly divisible by p ..
So if we chk for 9999(4 times),then it shud be exactly divisible by 5..but it is not the case...
Is there any mistake?
p shud be greater than 5:)
frm 7 onwards it works:)
viveknitw Sayswat is the last non zero digit of 25! and 35!
last digit of 25! can be cal. as follow.
the last digit of 4^n(2n!)
where n = N/20
for N= 20! n=2
so last digit of 20! = 4^2*4! = 4
so 25! = 5*4*3*2*1*20! = 8
now i guess u can col for 35! in the same manner:)
last digit of 25! can be cal. as follow.
the last digit of 4^n(2n!)
where n = N/20
for N= 20! n=2
so last digit of 20! = 4^2*4! = 4
so 25! = 5*4*3*2*1*20! = 8
now i guess u can col for 35! in the same manner:)
check the bold part once, take the full numbers not just the last digits
25! = 25*24*23*22*21*20! = 4
check the bold part once, take the full numbers not just the last digits
25! = 25*24*23*22*21*20! = 4
this is correct!
v cant avoid the reality that 25= 5*5
thus 4 is correct
last digit of 25! can be cal. as follow.
the last digit of 4^n(2n!)
where n = N/20
for N= 20! n=2
so last digit of 20! = 4^2*4! = 4
so 25! = 5*4*3*2*1*20! = 8
now i guess u can col for 35! in the same manner:)
the above bold part is wrong!
cant leave 5 o 25.
though the method will be good wen doin for 35 !
Z(30) = L(4^3)*Z(6)
Z(6) = 2 and L(4^3) = 4
so Z(30)= L(4*2) = 8
so z(35)= z (30) *1*2*3*4*5=8*1*2*3*4*5=6
m i wrong somwhere?