9. Convert each fraction to a terminating or repeating decimal:
1. 9/ 22
   solution:
2. 21/ 2048
   solution: .01025390625
3. 13/ 28
   solution:


10. Convert each repeating decimal to a quotient of integers and reduce to lowest terms. (Recall that 999 = 111·9 = 27·37.)
1. 
   solution:
    
   Hence:
2. 
   solution:
    
   Hence:
3. 
   solution:
    
   Hence:


11. Make a list of the distinct congruence classes modulo 9, in which each class is listed exactly once.
    solution: According to the standard representation, thee congruence classes are:
    [0]₉, [1]₉, [2]₉, [3]₉, [4]₉, [5]₉, [6]₉, [7]₉, [8]₉,.
    We can also name them as follows:
    [1]₉, [2]₉, [3]₉, [4]₉, [5]₉, [6]₉, [7]₉, [8]₉, [9]₉ .
     
12. Find the standard representation for each of the following congruence classes:
1. [71]₁₂.
   solution: 71 = 5·12 + 11, so that [71]₁₂ = [11]₁₂ .
2. [81]₉
   solution: 81 = 9·9, so that [81]₉ = [0]₉ .
3. [- 5]₁₃
   solution: [ - 5]₁₃ = [ - 5 + 13]₁₃ = [8]₁₃.
4. [- 13] ₅₁₃
   solution: [ - 13]₅ = [ - 13 + 3·5]₅ = [2]₅.
    
13. Find the standard class representative for each calculation:
1. 11 + 9 mod 12
   solution: or [11 + 9]₁₂ = [20]₁₂ = [8]₁₂ .
2. 11 · 9 mod 12
   solution: or [11 · 9]₁₂ = [99]₁₂ = [3]₁₂ .
3. 9 · 4 mod 12
   solution: or [9 · 4]₁₂ = [36]₁₂ = [0]₁₂ .
4. 4 - 7 mod 8
   solution: or [4 - 7]₈ = [-3]₈ = [5]₈ .
    
14. This year, Christmas will be on a Saturday - day 7 of the week, if we start the numbering with Sunday as day 1.
    On what day of the week will Christmas be next year? In the year 2001? Show appropriate mod 7 calculations to support your answer.
    Suggestion: The fact that 2000 is a leap year while 2001 is not should be taken into account.
    solution: 366 = 52·7 + 2, so that
    Therefore . So, Christmas will be on a Monday next year.
    To proceed to the year 2001, note that 365 = 52·7 + 1, so that .
    So, , and Christmas will be on a Tuesday in 2001.
     
15. Which of the following congruence classes have multiplicative inverses? Find the inverses of those that have inverses, and explain why the others don't have inverses.
1. [5]₈
   solution: [5]₈ · [5]₈ = [ 25]₈ = [1]₈. Thus, [5]₈ is its own inverse.
2. [5]₁₀
   solution: [5]₁₀ does not have an inverse, since 5 and 10 have a nontrivial GCD.
3. [4]₇
   solution: [4]₇ · [2]₇ = [8]₇ = [1]₇. Thus, [2]₇ is the inverse of [4]₇.
4. [4] ₁₀
   solution: [4]₁₀ does not have an inverse, since 4 and 10 have a nontrivial GCD.
    
16. Refer to the calculations from problem 3 to do the following:
1. Write GCD(16,37) as 16a+ 37b
   solution: We rearrange the two equations to obtain:
   5 = 37 - 2·16
   1 = 16 - 3·5
   Now, we substitute form the first equation for the red 5 in the second equation:
   1 = 16 - 3·(37 - 2·16)
   1 = 16 - 3·37 + 6·16
   So, finally:
   1 = 7·16 - 3·37
    
2. Write GCD(28,109) as 28a+ 109b
   solution: We rearrange the three equations to obtain:
   25 = 109 - 3·28
   3 = 28 - 25
   1 = 25 - 8·3 + 1.
   Now we substitute from the second equation for the red 3 in the third equation:
   1 = 25 - 8·(28 - 25)
   1 = 9·25 - 8·28
   Next, we substitute from the very first equation for the red 25 in our last equation:
   1 = 9(109 - 3·28) - 8·28
   1 = 9·109 - 27·28 - 8·28
   1 = 9·109 - 35·28
    
17. Use the answer from problem 16 to find multiplicative inverses of:
1. [16]₃₇
   solution: The equation from 16a) gives:
   [1]₃₇ = [7]₃₇·[16]₃₇ - [3]₃₇·[37]₃₇ .
   Since [37]₃₇ = [0]₃₇ , we conclude that [1]₃₇ = [7]₃₇·[16]₃₇ .
   Thus, [7]₃₇ is the multiplicative inverse of[16]₃₇ .
    
2. [28]₁₀₉
   solution: The equation from 16b) gives:
   [1]₁₀₉ = [9]₁₀₉·[109]₁₀₉ - [35]₁₀₉·[28]₁₀₉ .
   Since [109]₁₀₉ = [0]₁₀₉ , we conclude that [1]₁₀₉ = [ - 35]₁₀₉·[28]₁₀₉ .
   Thus, [ - 35]₁₀₉ is the multiplicative inverse of[28]₁₀₉ .
   We also can write the answer as [ - 35 + 109]₁₀₉ = [74]₁₀₉

Back to the review questions.

For questions: roberts@math.umn.edu