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1. Find 497 mod 101 using the fast exponentiation algorithm ...

1. Find 497 mod 101 using the fast exponentiation algorithm as given in lectures. 2. Find 4198 mod 101. 4. For the following values of (n, a) what result does the Miller-Rabin algorithm produce: (d) n = 21, a = 5 (e) n = 49, a = 19 (f) n = 39, a = 5 6. Alice wishes to send the message m = 4 to...

You may use any computer algebra system, but you may not use...

You may use any computer algebra system, but you may not use any functions on it other than powers, mods, a function to factor p - 1 and a function to compute Bezout's identity, if available. You must find the answer that way and present your reasoning. 1. Let p = 178481. (a) What is the order ...

2. Let ak be an arithmetic sequence. Recall that ak = a1 +(k...

2. Let ak be an arithmetic sequence. Recall that ak = a1 +(k-1)d", where d is the common difference between the terms. 71 a) Derive the formula Sn = ak = n(a, + an) using and properties of sums. 2 b) Verify the formula using and induction. 3. Use the strong form of induction to pro...

We say that a divisor d of n is 'special' if d+1 is also a d...

We say that a divisor d of n is 'special' if d+1 is also a divisor of n. Prove that at most half of the positive divisors of n are special, and find all positive integers n such that exactly half of the positive divisors of n are special.

Question 2: A function f is defined for all real numbers and...

Question 2: A function f is defined for all real numbers and satisfies f(2 + x) = f(2 - x) and f(7 + x) = f(7 - x) for all real x. If x = 0 is a root of f (x) = 0, what is the least number of roots of f F(x) = 0 must have in the interval - 1000 < x < 1000?

1. Determine whether the congruence below have any solutions...

1. Determine whether the congruence below have any solutions. If so, find all the solutions. [20 pts] 8x²+2xci= 0 (mod 13) 1009 2. Find (a) (b) ( 143 [20 pts] 2307 8243 3. Find a primitive root modulo 13. Express all possible primitive roots of 13 in terms of the one you found. Fin...

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