This exercise investigates what happens if we drop the assumption that gcd(e, p − 1) = 1 in Proposition 3.2. So let p be a prime, let c ≡ 0 (mod p), let e ≥ 1, and consider the congruence xe ≡ c (mod p). (3.36) (a) Prove that if (3.36) has one solution, then it has exactly gcd(e, p − 1) distinct solutions. (Hint. Use primitive root theorem (Theorem 1.30), combined with the extended Euclidean algorithm (Theorem 1.11) or Exercise 1.27.) (b) For how many non-zero values of c (mod p) does the congruence (3.36) have a solution?

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places

there would only be 50 odd numbers between 1 and 100 so therefor there would be 50 even so that would only be 50 odd numbers.

idk

step-by-step explanation: