what is the remainder when 2^311 is divided by 19. Show full detailed answer.
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2
Cool Question buddy
So we are going to use this trick formula
(a * b) %m = ((a%m)*(b%m))%m
so first base 2^10 mod 19 = 1024 mod 19 = 17
then 2^20 mod 19 = 17^2 mod 19 = 4
now 2^310 mod 19 = (17^31) mod 19 = (17^2)^15 * 17 mod 19
= 4^15 * 17 mod 9
now 4^5 mod 19 = 17
so 4^15 * 17 mod 9 = (4^5)^3 17 mod 19 = 17^4 mod 19
= 16
so 2^310 mod 19 = 16
2^311 mod 19 = 16 * 2 mod 19 = 13
So we are going to use this trick formula
(a * b) %m = ((a%m)*(b%m))%m
so first base 2^10 mod 19 = 1024 mod 19 = 17
then 2^20 mod 19 = 17^2 mod 19 = 4
now 2^310 mod 19 = (17^31) mod 19 = (17^2)^15 * 17 mod 19
= 4^15 * 17 mod 9
now 4^5 mod 19 = 17
so 4^15 * 17 mod 9 = (4^5)^3 17 mod 19 = 17^4 mod 19
= 16
so 2^310 mod 19 = 16
2^311 mod 19 = 16 * 2 mod 19 = 13
Anonymous286:
congruence modulo method
Answered by
2
Answer:
Cool Question buddy
So we are going to use this trick formula
(a * b) %m = ((a%m)*(b%m))%m
so first base 2^10 mod 19 = 1024 mod 19 = 17
then 2^20 mod 19 = 17^2 mod 19 = 4
now 2^310 mod 19 = (17^31) mod 19 = (17^2)^15 * 17 mod 19
= 4^15 * 17 mod 9
now 4^5 mod 19 = 17
so 4^15 * 17 mod 9 = (4^5)^3 17 mod 19 = 17^4 mod 19
= 16
so 2^310 mod 19 = 16
2^311 mod 19 = 16 * 2 mod 19 = 13
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