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# MAT135-Assignment5-GradingScheme - MATH 135 Fall 2010...

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Unformatted text preview: MATH 135, Fall 2010 Solution of Assignment #5 Problem 1. Find the prime factorization of each of the following integers. (a) 20!. (b) (25°) SOLUTION. (a) First, let us describe a method for ﬁnding the exponent of a prime p in the prime factoriza- @ tion of n! for any positive integer 71. Note that n! is the product of the numbers 1, 2, 3, . 1 - ,n. The multiples of p that occur in this list are 11,210,310, - - - , I%I’ so there are Igj multiples 71 of p in the list. Similarly, there are IFZI multiples of p2 in the list and Iﬁj multiples of p3 and so on. Thus the exponent of the prime p in the prime factorization of n! is equal to Iﬁl+l§zl+ [eh-~- Using the above rule, the exponent of21n 20! isw + 2—0 + 29 = 10+5+2+1— _ V2 18, the exponent of 3 IS I29 3 + 33—20 — ”6+2 — 8, t e exponW, the exponent V), [email protected] of 7 is I2—70I = 2, and the exponents of 11, 13, 17, and 19 are all equal to 1. Thus we have 2. @562) M W 20!221838-5431?1113.17.19. @ /———~——/ (b) Using our rule from part (a) we ﬁnd that 5!:22+1-3.5,and 136) 15! = 27f3+1.35+1-53172.11~13 and so 20 20! 218-38.54.72.11~13-17-19 <10>——~—-———————————-—- @ 5115! (23 . 3 - 5)(211 - 36 - 53 - 72 . 11 - 13) =24.3-17-19| Problem 2. Determine all n 6 11” for which In2 + 2n — 63I is a prime number. Be sure to justify @ your answer. e> @ SOLUTION. Note that 712+ 271—35 :In1—7Liﬁﬂjjn —- 7 71 +9 F.or| In2 + 2n — 63I to be prime, then either In — 7— — 1 or n + 9: 1(since otherwise I112 + 2n — 63] would be factored as1®+ "[email protected] the produc 0 two non—negative integers, neither of which' 1s 1).€1<(>\sm\lw IfIn—7I—-1thenn—7—i1son=80rn=6. }@ If In+9I =1, thenn+9=i1, son=—80rn=—10. W. This does not guarantee that I712 + 2n — 63I is prime though, GD as we have to check the other factor. If n— — 8, then In2 + 2n — 63I: I64 + 16~ 63I— — 17, which [email protected] me. If n— — 6, then I712 + 2n — 63I: I36 + 12— 63L: 15 which is not rime. To summarize, if @ In2 + 2n —_63| is prime, then 11— —~ 8 or 77,— -* 6. However, In2 + 271— 63I IS actually prime only when n— — 8. l 2 Problem 3. Suppose that p, q and r are prime numbers and that p is odd. If p I 2g + r and p I 2q — 7‘, prove that q = 7". SOLUTION. Since p I 2q +r and p I 2q — 7', then p I (2q+ r) + (2q — r) 2 4q by Proposition 2.11.ii. @ Also, since p I 2q + 7‘ and p I 2q — r, then p I (2g + 7‘) — (2q — r) = 27‘ by Proposition 2.11.ii. @ Since 1) is prime and p I 4q, then p I 4 or p I q. Since p is odd and prime, we cannot have p I 4, so p I q. Since q is prime, p is prime and p I q, then q = p (because the only divisors of q are 1 and q). @ Since p is prime and p I 2r, then p I 2 or p I r. 6/) Since 19 is odd and prime, we cannot have p I 2, so p I 7‘. @ Since q is prime, p is prime and p I r, then 7" = p. Sincequandr2p,thenqzrzp. I ® Problem 4. Show that for all positive integers a and b we have aIb if and only if a2Ib2. SOLUTION. Write a =Ap1j1p2j2 - - 107/" and b = 101’“ka2 - - 497,19" where the p, are distinct primes. :I (Q Then we have a2 = p12J1p22J2 . . .pn2Jn and b2 = p12k1p22k2 - - 491,216”, and so 3 CD aIb 4:» j, g k,- for allz‘ <==> 23'.- szk, for alli 4:» a2Ib2. I 3(3) ...
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