Question

how much is (7^100240) mod 13? how much is (7^100240) mod 13? @Mathematics

Answer 1

7^100240=49^50120=10^50120(mod 13)=1000^(50120/3) (mod 13)= (-1) ^ (50120/3)=(-1)(2/3)=1

Answer 2

is it right that 49^50120=10^50120(mod13)? because 49mod 13=10, but 49^2 mod 13=9??

Answer 3

9(mod13)=100(mod 13)=10^2 (mod 13)
got it? :)

Answer 4

hello?? can u understand?

Answer 5

not really

Answer 6

which part?

Answer 7

have u read divisibility??

Answer 8

yes, some kind of

Answer 9

good, then u must know 100 (mod13)=9(mod 13), as when 100 is divided by 13, the remainder will be 9

Answer 10

yes

Answer 11

then i cant get what ur real problem is :(

Answer 12

i have solved it in easy method

Answer 13

oh, i guess i understood it, just took some time

Answer 14

thanks :)

Answer 15

one last question- is 50120/3=2/3 because 50120 mod 3=2?

Answer 16

\[7 ^{100240}\mod13\]you ca use the Carmichael function (lambda function) http://mathworld.wolfram.com/CarmichaelFunction.html and http://en.wikipedia.org/wiki/Carmichael_function
\[100240=2 ^{4}*5*7*179\]\[\lambda(100240)=LCM(\lambda(2),\lambda(5),\lambda(7),\lambda(179))=LCM(1,4,6,178)=1068\]this theorem then states that\[7 ^{1068} = 1\mod13\]so now you can take off all multiples of 1068 from 100240 leaving 916 - so\[7 ^{100240}\mod13=7 ^{916}\mod13\]you can then re-apply the Carmichael function:\[916=2 ^{2}*229\]\[\lambda(916)=LCM(\lambda(2),\lambda(229))=LCM(1,228)=228\]so now take off all multiples of 228 from 916 leaving 4 - so\[7 ^{916}\mod13=7 ^{4}\mod13=2401\mod13=9\]