Approximate the sum of the series correct to four decimal places. į (-1)"n 81 n=1 S
Suppose a
c mod n and bd
mod n.
(a) show that a + b
c + d mod n
(b) show that a * b
c * d mod n.
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4. Show that the following congruence is true ab = (a mod n)" (mod n) for any positive integers a, b, and n.
Problem 4. Show that for all integers n, n2 mod 3- 1 n2 mod 3- 0 or (i.e. there exists an integer k such that n2 3k or n2 3k +1). mp
Let m be a positive integer. Show that a mod m - b mod m t a - b (mod m) Drag the necessary statements and drop them into the appropriate blank to build your proof (mod m Dag the mecesary eemnes a ohem int the approprite Proof method: Proof assumptions), at-qm + Proof by contradiction aaandh mam it Implication(s) and deduction(s) resulting from the assumption(s): a mk + bmk Hqm tr a-(k + q)m+ r Conclusion(s) from implications and...
(1) Show that the non-zero residue classes of the integers (mod n) form a group under multiplication if n is prime. motional numbers, let addition and
A. Find the multiplicative inverse of 52 mod 77. Your answer should be an integer s in the range from 0 through 76. Check your solution by verifying that 52s mod n = 1. Show that for all integers a, b, and c, if aſb and alc, then a-|bc.
2. Find 11644 mod 645 Use the following algorithm and show work! procedure modularExponentiation(b: integer, n = (ak-1ak-2...a1a0)2, m:positive integer) x:= 1 power := b mod m for i = 0 to k-1 If ai = 1 then x:= (x⋅power) mod m power := (power⋅power) mod m return x ( x equals bn mod m) Note: in this example m = 645, ai is the binary expansion of 644, b is 11.
Problem 6: There are three users with pairwise relatively prime moduli n, n and n3. Suppose that their encryption exponents are all e3. The same message m is sent to each of them and you intercept the three ciphertexts ci mrs (mod n.), for i-1, 2, 3. (a) Show that 0 m3< nin2n (b) Show how to use the CRT to find m3 (as an exact integer, not only as m3 (mod ninns)) and, therefore also m c) Suppose that...
1. Show that a 1728 = 1 (mod p) when p= 7, 13, 19 for all a E N such that p /a. 2. Let p be a prime and p = 3 (mod 4). Show that r2 = -1 (mod p) has no solution. (Hint: Raise both side to (p-1)/2.)