Use Fermat’s theorem to find a number x between 0 and 37 with x^73 congruent to 4 modulo 37. (You should not need to use any brute-force searching.)
Fermat's Little Theorem states that if "p" is a prime number and "a" is an integer not divisible by "p," then:
a(p-1) ≡ 1 (mod p)
Let's use Fermat's theorem to find a number "x" such that x73 ≡ 4 (mod 37):
Since 37 is a prime number, we can use Fermat's Little Theorem with "a = x" and "p = 37":
x(37-1) ≡ 1 (mod 37)
x36 ≡ 1 (mod 37)
Now, rearrange the congruence equation to fit our requirement:
x73 = x(2 * 36 + 1)
Using the property of modular arithmetic:
x(2 * 36 + 1) ≡ (x36)2 * x ≡ 12 * x ≡ x (mod 37)
So, need to find a number "x" such that "x" is congruent to 4 (mod 37). A straightforward solution is x = 4. Let's verify:
473 ≡ 4 (mod 37)
Now, let's calculate (473) % 37:
(473) % 37 = 4
Therefore, x = 4 is the number between 0 and 37 that satisfies x73 ≡ 4 (mod 37) using Fermat's theorem.