Test 3 Solution - MAS 4214/001 Elementary Number Theory...

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MAS 4214/001 Elementary Number Theory, Fall 2013TEST 3Name:Date: 12/04/2013(2 points)Time allowed: 50 minutesShowALLsteps. Fifty points equal 100%.Assume in this test that all small Roman letters represent integers, andm1.Question 1.(5 points)True or false: Explanation is not needed.(a) Ifn4 is composite, then (n1)!negationslash≡ −1 (modn).(b) If gcd(a,m) = 1, then an inversea, modulom, ofaexists and gcd(a,m) = 1.(c) The following congruences are equivalent:11x6(mod 24),(9)(11x)(9)(6)(mod 24),3x6(mod 24), x2(mod 8).(d) Ifmis composite andbm11 (modm) for someb2, thenmis a psuedoprime to the baseb.(e) If 5m12 (modm), thenmis composite.Answer:
Question 2.(2+6 points)Consider the congruence 5513xc(mod 13559). It is known thatgcd(13559,5513) = 149and149 = (13559)(13) + (5513)(32).(a) For which values ofcdoes the congruence have an solution forx?
(b) Solve the congruence 5513x1639 (mod 13559). Note that 1639 = (149)(11).

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