University:
Amherst College
Course:
MATH 450 | Measure Theory and Integration
Academic year:

2015
Views:

433

Pages:

8
Author:

Rafael Werner

MATH 42 MIDTERM 1 20 FEBRUARY 2015 Name: Solutions The time limit is 50 minutes. No calculators or notes are permitted. Each problem is worth 5 points. (1) Find all prime numbers p between 1 and 100 such that p-1 (mod 15). The numbers congruent to Imod 15 lectween I and 100 are: 14,29, 44.59,74,89 Of these. 3 are even. But the rest turn out to be prime. 29.59.89 (2) Recall that a primitive Pythagorean triple consists of three pos- itive integers (a, b, c) such that a²+b²=c, and there are no common factors of a, b and c. Find a primitive Pythagorean triple such that a 15. We wish to solve 15²+b²=c² ie. 15²= (c+b)(c-b). Since c.b should have no common fador, neither should ctb, c-b. Therefore there are two ways to find solutions: Sc+b = 5² 1c-6=32 => [c=2+2=17 (a.b.c)= (15,8,17) C+b=157 c-b=1 => = 1322 113 b=112 (a,b,c) = (15,112,113) Both are PPTr, and in fact thekare the only possible choices Remark Both can be found easily by the recipe (st. S't sit). The logiz above follows the logic of how this recipewas found. (3) Compute the greatest common divisor of 1106 and 203. 1106-5-203 1106-1015 =91 =21 203-2-91203-182 91-4-2191-84 21-3-7=0. =7 Hence qud (1106.203) = gcd(203.91) = ged (91.21) = gcd (21.7) =qid 17.0)=7. (4) Solve the following congruence. 2823 (mod 149) Applying the (extended) Euclidean algorithm to 28 and 149: (149) (28) [9] = (149)-5-(23) [1] = (28)-3-[9] = (28)-3-[(149)-5-(28)] = 16-(28)-3(149). Thus 16-281 mod 149. Therefore 28x = 3 (mod149) 16.28x 16.3 mod 149 x = 48 mod 149 (5) Suppose that a, b, c are positive integers such that ged(a, b) = 1. Prove that if a divides be, then a divides c. Solution 1 Since gel(a,b)=1, there exist xy E72 st. ax+by=1. Then cax +cby=c a. (cx+y)=c Since alba, Z. Su cx+y EZ, so cis a midtipleoda. Solution 2 Make use of unique Pactorization into primes. Then a,b,c can be written a = Pipe P } each Pi. 4. Ti prime. b = 9.99m (= ΓιΓεη (not necessarily distinct). Then the albe meam that ; it too can be factored into primes. By uniquenes this factorization includes qam. r., in except that each p... pe has been removed thom the list. Since ged(a,b)=1, no prime pi equads a prime ai; so in Pact each pi occions (the right numberol times) in the factorization of a alone. Hence alc.

Math 42 Midterm 1 2015

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