psol03

# psol03 - Week 3 Homework Soultions Chapter 2 Section 4...

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Week 3 Homework Soultions Chapter 2: Section 4: 12.39 = 3(13) 81 = 3 4 101 is prime 143 = 11(13) 289 = 17 2 899 = 29(31) 14.A number is divisible by 10 n iff it ends with n zeroes. Hence we wish to find the greatest power of 10 which divides 100!. Clearly 2 divides 100! more than 5 does, so we wish to find the greatest power of 5 which divides 100!. We observe that 25, 50, 75, and 100 each provide two factors of five and 5, 10, 15, 20, 30, 35, 40, 45, 55, 60, 65, 70, 80, 85, 90, and 95 each provide one factor of five. Hence 100! contains 24 factors of five so the number 100! ends with 24 zeroes. 16.1, 5, 7, 11 20.A. 1 + 2 + 3 = 6 1 + 2 + 4 + 7 + 14 = 28 B. When 2 p -1 is prime the factors of 2 p-1 (2 p -1) are 1, 2, 4, . . . , 2 p-1 and 1(2 p -1), 2(2 p -1), 4(2 p -1), . . . , 2 p-1 (2 p -1). The sum of the first row is 2 p-1 and the sum of the second row is 2 p -1 times that of the first. Hence the sum of all the factors is 2 p-1 (2 p -1+1) = 2 p (2 p-1 ). Subtracting the number 2 p-1 (2 p -1) itself gives us 2

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psol03 - Week 3 Homework Soultions Chapter 2 Section 4...

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