Question

Use Fermat’s Little Theorem to compute 3302 mod 5, 3302 mod 7, and 3302 mod 11.

a) 2, 6, 9
b) 2, 1, 2
c) 2, 4, 7
d) 2, 5, 10

Answer 1

To compute 3302 mod 5, 3302 mod 7, and 3302 mod 11 using Fermat's Little Theorem, we can use the formula a^(p-1) mod p = 1.

Step-by-step explanation:

Fermat's Little Theorem states that if p is a prime number and a is not divisible by p, then a^(p-1) mod p = 1. To calculate 3302 mod 5, we can use this theorem because 5 is a prime number and 3302 is not divisible by 5. Therefore, 3302^(5-1) mod 5 = 1. Simplifying the exponent, we have 3302^4 mod 5 = 1. Since 3302 = (5 * 660) + 2, we can rewrite the expression as 2^4 mod 5 = 1. Evaluating 2^4 mod 5, we get 16 mod 5 = 1, which means that 3302 mod 5 is equal to 1.

To calculate 3302 mod 7, we can again use Fermat's Little Theorem because 7 is a prime number and 3302 is not divisible by 7. Therefore, 3302^(7-1) mod 7 = 1. Simplifying the exponent, we have 3302^6 mod 7 = 1. Since 3302 = (7 * 471) + 5, we can rewrite the expression as 5^6 mod 7 = 1. Evaluating 5^6 mod 7, we get 15625 mod 7 = 1, which means that 3302 mod 7 is equal to 1.

To calculate 3302 mod 11, we can once again use Fermat's Little Theorem because 11 is a prime number and 3302 is not divisible by 11. Therefore, 3302^(11-1) mod 11 = 1. Simplifying the exponent, we have 3302^10 mod 11 = 1. Since 3302 = (11 * 300) + 2, we can rewrite the expression as 2^10 mod 11 = 1. Evaluating 2^10 mod 11, we get 1024 mod 11 = 1, which means that 3302 mod 11 is equal to 1.