Math 312. Assignment 8 solutions.
Problem 7.2.4 Let n = 2d pd1 . . . pdr , where d 0, d1 , . . . , dr 1 and p1 , . . . , pr
are odd primes. Then
(n) = (2d )(pd1 ) . . . (pdr )
is odd, if and only if each factor is odd. Recall that
(pr ) = 1 + p +
Math 312. Assignment 6 solutions.
Problem 5.1.4. Since 5 26 (mod 31), 4 27 (mod 31), etc., we have
5! 25! 25! (26) (27) (28) (29) (30) (1)5 30!
By Wilsont theorem, the last expression equals (1) (1) 1 (mod 31).
Problem 5.1.10. By Fermats theore
COMMENTS ON ASSIGNMENT 1
Handwriting. Many papers are hard to read because of bad handwriting.
Please try to write clearly and keep your notations consistent.
Explain your solutions. It is not enough to write down formulas on the page;
Math 312. Assignment 5 solution outlines.
Problem 4.3.2. Need to solve the system of congruences
x 1 (mod 2)
x 0 (mod 3)
x 1 (mod 5)
for x. Note that 2, 3 and 5 are pairwise relatively prime. The formula in the book
x 1 M1 y1 + +0 M2 y2 + 1 M3 y3 (
COMMENTS ON ASSIGNMENT 2
General: Once again, presentation is important. Please solve the problem and
clean up your solution before writing it down. Indicate intermediate steps and
leave out observations and arguments that are not needed for the solution.
COMMENTS ON ASSIGNMENT 3
The marker tell me that some of the HW3 papers are virtually identical to
each other, down to the notation and the wording. We assumed this resulted from
misunderstanding of the rules (posted on the course website
Math 312. Assignment 9 solutions.
Problem 8.1.2 Answer: I CAME I SAW I CONQUERED
Problem 8.1.4 Answer: NPWJEAPNSPQESW
Problem 8.1.6 Solving C 3P + 24 (mod 26) for P , we obtain 3P C + 2
P 31 (C + 2) 9(C + 2) 9C + 18 9C 8
(mod 26) .
Using this for
Math 312. Assignment 10 solutions.
Problem 8.4.8 To break the code, we factor 2881 = 43 67. Now (2881) =
42 66 = 2772. Using the Euclidean algorithm, we compute the decryption key
d e1 51 1109 (mod 2772).
We now break up the ciphertext into blocks of four
Math 312. Assignment 7 solutions.
Problem 6.2.18(a) Let n = p1 p2 p3 , where p1 = 6m + 1, p2 = 12m + 1 and
p3 = 18m + 1 are primes. We have to show that if gcd(a, n) = 1, then an1 1
(mod n). By the Chinese Remainder Theorem it suces to show that
an1 1 (mo
COMMENTS ON ASSIGNMENT 4
1. Comments from the marker
Some solutions still look quite similar to each other. Remember, that you may
talk about the problems together but you are responsible for writing out your own
solution, in your own words.
From now on,