Again with the vague theorem referrence. This time you clearly mean Eulers Totient Theorem. Why not reduce once more?pivoxa15 said:I have got another question, this time involving the Euler's Theorem:
a^(phi(m)) is congruent to 1 (mod m)
The question is calculate
7^40002 mod 1000
I could only reduce it to
7^402 mod 1000
What should I do now?
Thanks
Euler's Theorem is a mathematical theorem that states that if two numbers, a and m, are relatively prime (meaning they have no common factors other than 1), then a raised to the phi function of m, where phi is Euler's totient function, is congruent to 1 modulo m.
Euler's Theorem is used to simplify the calculation of 7^402 mod 1000 by reducing the exponent from 402 to a smaller number. This is done by finding the phi function of 1000, which is 400, and using it as the new exponent in the equation, resulting in 7^400 mod 1000.
The phi function of 1000 is the number of positive integers less than or equal to 1000 that are relatively prime to 1000. In this case, the phi function of 1000 is 400, as the only positive integers less than or equal to 1000 that are relatively prime to 1000 are 1, 3, 7, and their multiples up to 1000.
To calculate 7^400 mod 1000, we can use Euler's Theorem to reduce the exponent to a smaller number. Since 7 and 1000 are relatively prime, we can replace 402 with the phi function of 1000, which is 400. Then, we can use the property of modular arithmetic that states (a^b mod m) = (a^b mod m) * (a^b mod m) mod m. This results in 7^400 mod 1000 = (7^4 mod 1000) * (7^4 mod 1000) * (7^4 mod 1000) * (7^4 mod 1000) mod 1000. We can then calculate each term separately and take the product of the results to get the final answer.
The final result of the calculation is 1. This is because Euler's Theorem states that 7^400 mod 1000 is congruent to 1 modulo 1000. Therefore, 7^402 mod 1000 is equivalent to (7^400 mod 1000) * (7^2 mod 1000) mod 1000, which is equivalent to (1 * 49) mod 1000, which gives us a final result of 49.