Math 3113, Section 4
Fall 1999
Solutions to chapter 5 and 6 review questions
(Problems 9 through 17)
 Convert each fraction to a terminating or repeating decimal:
 9/ 22
solution:
 21/ 2048
solution: .01025390625
 13/ 28
solution:
 Convert each repeating decimal to a quotient of integers and reduce to lowest terms. (Recall that 999 = 111·9 = 27·37.)

solution:
Hence:

solution:
Hence:

solution:
Hence:
 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} .
 Find the standard representation for each of the following congruence classes:
 [71]_{12}.
solution: 71 = 5·12 + 11, so that [71]_{12} = [11]_{12} .
 [81]_{9}
solution: 81 = 9·9, so that [81]_{9} = [0]_{9} .
 [ 5]_{13}
solution: [  5]_{13} = [  5 + 13]_{13} = [8]_{13}.
 [ 13]_{ 513}
solution: [  13]_{5} = [  13 + 3·5]_{5} = [2]_{5}.
 Find the standard class representative for each calculation:
 11 + 9 mod 12
solution: or [11 + 9]_{12} = [20]_{12} = [8]_{12 }.
 11 · 9 mod 12
solution: or [11 · 9]_{12} = [99]_{12} = [3]_{12 }.
 9 · 4 mod 12
solution: or [9 · 4]_{12} = [36]_{12} = [0]_{12 }.
 4  7 mod 8
solution:
or [4  7]_{8} = [3]_{8} = [5]_{8 }.
 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.
 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.
 [5]_{8}
solution: [5]_{8} · [5]_{8} = [ 25]_{8} = [1]_{8}. Thus, [5]_{8} is its own inverse.
 [5]_{10}
solution: [5]_{10} does not have an inverse, since 5 and 10 have a nontrivial GCD.
 [4]_{7}
solution: [4]_{7} · [2]_{7} = [8]_{7} = [1]_{7}. Thus, [2]_{7} is the inverse of [4]_{7}.
 [4]_{ 10}
solution: [4]_{10} does not have an inverse, since 4 and 10 have a nontrivial GCD.
 Refer to the calculations from problem 3 to do the following:
 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
 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
 Use the answer from problem 16 to find multiplicative inverses of:
 [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} .
 [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