55SS3 - SOLUTION SET #3 FOR MATH 55 Note. Any typos or...

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SOLUTION SET #3 FOR MATH 55 Note. Any typos or errors in this solution set should be reported to the GSI at isammis@math.berkeley.edu 2.4.10. What are the quotient and remainder when (a) 19 is divided by 7? (b) -111 is divided by 11? (c) 789 is divided by 23? (d) 1001 is divided by 13? (e) 0 is divided by 19? (f) 3 is divided by 5? (g) -1 is divided by 3? (h) 4 is divided by 1? Solution. The key thing for this entire problem is to remember the definition of the division algorithm, which requires the remainder r of a divided by b to be in the range 0 r < b . So: (a) Quotient 2, Reminder 5: 19 = 2(7) + 5. (b) Quotient -11, Remainder 10: - 111 = - 11(11) + 10 (c) Quotient 34, Remainder 7: 789 = 34(23) + 7 (d) Quotient 77, Remainder 0: 1001 = 77(13) + 0 (e) Quotient 0, Remainder 0: 0 = 0(19) + 0 (f) Quotient 0, Remainder 3: 3 = 0(5) + 3 (g) Quotient -1, Remainder 2: - 1 = - 1(3) + 2 (h) Quotient 4, Remainder 0: 4 = 4(1) + 0 2.4.14. How many zeros are there at the end of 100!? Solution. A zero at the end of the factorial represents a 10 divisor. Since twos appear in the prime factorization of 100! far more frequently than do fives, we can simply count the number of factors of 5 that must appear in 100!’s prime factorization. First, every multiple of five contributes one factor of five to 100!. Since there are 100 / 5 = 20 mutliples of five in 100!, we have twenty factors of five. Now, each multiple of 25 contributes an additional factor of five, so we have four more, for a grand total of 24 factors of 5. Again, there are many more than 24 factors of two in 100! (there are 50 just from multiples of two, before even counting the additional factors from multiples of 4, 8, 16, and so forth). Thus, ever factor of five can be matched with a factor of two, and results in a zero at the end of 100!. 100! thus terminates in a string of 24 zeros. Typeset by A M S -T E X 1
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2 SOLUTION SET #3 FOR MATH 55 2.4.16. Which positive integers less than 12 are relatively prime to 12? Solution. Since 12 = 2 2 3 , the integers that aren’t multiples of two or three are relatively prime to 12. Those are { 1 , 5 , 7 , 11 } . 2.4.20. We call a positive integer perfect if it equals the sum of its positive divisors other than itself. (a) Show that 6 and 28 are perfect (b) Sow that 2 p - 1 (2 p - 1) is a perfect number when 2 p - 1 is prime. Solution. (a) The positive divisors of 6 (other than 6) are 1, 2, and 3. 1+2+3=6, so 6 is perfect. The positive divisors of 28 (other than 28 itself) are 1, 2, 4, 7, and 14. Since 14 + 7 + 4 + 2 + 1 = 28, 28 is also perfect. (b) Suppose 2 p - 1 is prime. Then the factors of 2 p - 1 (2 p - 1) other than the number itself are 2 j , where j = 0 , 1 ,... ,p - 1 and (2 p - 1)2 j for j = 0 , 1 ,... ,p - 2. Summing these, we see
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55SS3 - SOLUTION SET #3 FOR MATH 55 Note. Any typos or...

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