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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)
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20!2218385431?1113.17.19. @
/———~——/ (b) Using our rule from part (a) we ﬁnd that 5!:22+13.5,and 136) 15! = 27f3+1.35+153172.11~13 and so 20 20! 21838.54.72.11~131719
<10>——~————————————— @ 5115! (23 . 3  5)(211  36  53  72 . 11  13)
=24.31719 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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 Spring '08
 ANDREWCHILDS

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