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remainder of (16937^30)/31
how to solve these remainder questions??
Read Solution (Total 10)
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- a^(p-1)/p
Is always a remainder is 1
Where p is a prime no. - 10 years agoHelpfull: Yes(89) No(1)
- = ((16937%31)^30)%31 i.e (11^30)%31 (by remainder theorem)
= (121^15)%31
=(28^15)%31 =28(784^7)%31
=28(9^7)%31
=28*9*(19^3)%31
=1728468%31=1 - 10 years agoHelpfull: Yes(7) No(1)
- (16937^30)/31....
(( 546 X 31 ) + 11)^30)/31
= 11^30/31
=(11^6)^5/31
=(57147*31+4)^5/31
=4^5/31
1024/31
Remainder is 1 - 10 years agoHelpfull: Yes(3) No(0)
- powers of 7 always follows the unit place digit sequence as 7 , 9 , 3 , 1
7^1 ends with =7
7^2 ends with = 9
7^3 ends with = 3
7^4 ends with = 1
similarly
7^29, 7^30, 7^31, 7^32 ends with 7, 9, 3, 1 respectively
- 10 years agoHelpfull: Yes(1) No(13)
- 16937/31= r(11)
(11)^30 /31= it can be written as 1331/31=r(29)
29/31=r(-2)
2^1000/31=r(1)=>1024/31=r(1) - 10 years agoHelpfull: Yes(1) No(0)
- a^(p-1)/p
the remainder is 1
p is prime no. - 9 years agoHelpfull: Yes(1) No(0)
- a^(p-1)/p
Is always a remainder is 1
Where p and a shuld be prime no. - 9 years agoHelpfull: Yes(1) No(1)
- can you explain clarly the concept
- 10 years agoHelpfull: Yes(0) No(2)
- unit digit of (169337^30) =49
49/30=1.6 - 10 years agoHelpfull: Yes(0) No(6)
- simple
a^(p-1)/p
Is always a remainder is 1
Where p is a prime no.
- 9 years agoHelpfull: Yes(0) No(0)
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