MATH10101: problems for supervision in week 12 SOLUTIONS

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1 MATH10101: problems for supervision in week 12 SOLUTIONS Q37.. Use the table below to sieve the integers up to 200 for primes. Thus calculate π200. What is the smallest composite number greater than 200 that has no prime factor less than 200? Are the following numbers prime or composite: i 44517; ii 44503; iii 44519?

2 Q37 - solution. Primes up to 200: Thus π200 = 46. Suppose that a composite integer N has no prime factors in the table. Let p be the smallest prime factor of N; then p is not in the table i.e. p > 200. Check that the smallest prime not in the table is 211; hence p 211. Also recall that p N. We conclude that N p hence = is the smallest composite integer that has no prime factor in the table. i is composite: one can observe that is divisible by 3 e.g. the sum of its digits is = 21 a multiple of 3. Recall that a positive integer is congruent modulo 3 to the sum of the digits of its decimal expansion. Alternatively = = = The prime factorisation of is ii = I hope you started looking for prime factors near 200 rather than 2. There does not seem to be an obvious way to factorise this number but to look for a prime p p < 211 such that p

3 Another possible approach to showing that is composite is to use Fermat s Little Theorem: mod hence cannot be prime. Calculating this involves many steps of successive squaring how many? iii is prime. If composite it would have a prime factor i.e. less than 210 and thus be in the table. It should have taken you 46 trial divisions to find that none of the 46 primes divided and thus it was itself prime. Q38. i Find the four smallest sets of prime triplets of the form p p + 4 and p + 6. Hint Use the table from Q37 ii Why are there no prime triplets of the form p p + 2 and p + 4 other than 3 5 7? Hint look at the p modulo 3 Q38 - solution. i and ii We will show that if p 1 or 2 mod 3 then at least one of p p + 2 and p + 4 is a multiple of 3 and thus not prime. Consider three cases. a If p 1 mod 3 i.e. p = 1 + 3k then p + 2 = 31 + k is not prime. b If p 2 mod 3 i.e. p = 2 + 3k then p + 4 = 32 + k is not prime. c If p 0 mod 3 then the only prime divisible by 3 is 3. This leads to the given triplet of Q39. Show that Euler s phi function evaluated at prime powers satisfies φp k = p k 1 p 1. Hint Instead of counting the set of integers coprime to p k count the complement of this set. Q39 - solution. Recall that φn = {r : 0 r n 1 r is coprime to n}. Follow the hint and count the number of integers 0 r p k 1 that are not coprime with p k. Not coprime with p k means a common divisor with p k greater than 1. But all positive divisors of p k are powers of the prime p hence all such integers are of the form mp where 0 m pk 1 p = p k 1 1 p. But m is an integer so in fact 0 m p k 1 1. Thus there are p k 1 integers not coprime to p k out of a possible total of p k integers. Hence φp k = p k p k 1.

4 Q40. Use Fermat s Little Theorem and Euler s Theorem to i show that is divisible by 7 ii show that is divisible by 3 but not by 9 iii find the last two digits in the decimal expansion of iv Calculate the remainders of 7 5 mod 13 and 7 7 mod 13. v Using Fermat s Little Theorem along with part iv solve 6x 5 mod 13. Do not use Euclid s algorithm. Q40 - solution. i We have mod 7 e.g = mod 7. Note that 4 is coprime to 7 so by Fermat s Little Theorem mod 7. But 2222 = Thus mod 7. We also have mod 7 and 3 7 = 1 and so by Fermat s Little Theorem mod 7. But 5555 = So mod 7. Hence mod 7. Alternatively note that mod 7 so = Now mod 7 so = which is 1 mod 7 by Fermat s Little Theorem. Hence we obtain modulo = 0. ii We immediately examine mod 9 not mod 3. Because 9 is not prime we need to use Euler s Theorem which in this case says that if gcda 9 = 1 then a φ9 1 mod 9. By Q39 or simply by listing the integers less than 9 we see that φ9 = 6. We have mod 9 and mod 9. Recall that a positive integer is congruent modulo 9 to the sum of the digits of its decimal expansion. Thus mod 9 using that 5555 is odd. Euler s Theorem and 2222 = together give = 3 mod 9.

5 Hence mod 9. Thus is of the form 3 + 9l for some l Z. Here 3 + 9l = l is a multiple of 3 i.e. divisible by 3 but not by 9. iii First mod 100. Euler s Theorem modulo 100 states that if gcda 100 = 1 then a 40 1 mod 100 since φ100 = 40. Noting that 7777 = while 3333 = gives = mod 100 since = = 1. Finally using the method of successive squaring we get mod 100 and mod 100. Hence the last two digits of are 90. iv mod 13 and mod 13. v We notice that 6x 5 mod 13 is by part iv the same as 7 7 x 5 mod 13. Fermat s Little Theorem states that mod 13. So multiplying our congruence by 7 5 gives 7 12 x mod 13 i.e. x 3 mod 13. Q41. Using the method of successive squaring calculate 2 90 mod 91. Hence show that 91 is not prime. Q41 - solution. mod Note how we are lucky here to get a repeating pattern. Then 2 90 = mod 91.

6 If 91 were a prime then Fermat s Little Theorem gives since gcd2 91 = = mod 91. This contradiction means that 91 is not a prime. Q42. Let σ 1 = σ 2 = σ 3 = S 6 S 6 S 6 i Calculate σ 1 σ 2 σ 2 σ 3 σ 3 σ 1 σ 2 1 σ 3 3 σ 1 σ 2 σ 1 ii Find the inverses of σ 1 σ 2 σ 1 σ 2 σ 3. iii Verify that σ 1 σ 2 1 = 2 1 directly multiplying the permutations on the righthand side. Q42 - solution. i σ 1 σ 2 = σ 2 σ 3 = σ 3 σ 1 = σ1 2 = σ3 3 = σ 1 σ 2 σ 1 =

7 ii 1 = 2 = σ 1 σ 2 1 = 3 = iii 2 1 = = = σ 1 σ 2 1.