 ### Solutions to chapter 5 and 6 review questions

#### (Problems 9 through 17) 1. Convert each fraction to a terminating or repeating decimal:
1. 9/ 22
solution: 2. 21/ 2048
solution: .01025390625
3. 13/ 28
solution: 2. 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: 3. 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:
9, 9, 9, 9, 9, 9, 9, 9, 9,.
We can also name them as follows:
9, 9, 9, 9, 9, 9, 9, 9, 9 .

4. Find the standard representation for each of the following congruence classes:
1. 12.
solution: 71 = 5·12 + 11, so that 12 = 12 .
2. 9
solution: 81 = 9·9, so that 9 = 9 .
3. [- 5]13
solution: [ - 5]13 = [ - 5 + 13]13 = 13.
4. [- 13] 513
solution: [ - 13]5 = [ - 13 + 3·5]5 = 5.

5. Find the standard class representative for each calculation:
1. 11 + 9 mod 12
solution: or [11 + 9]12 = 12 = 12 .
2. 11 · 9 mod 12
solution: or [11 · 9]12 = 12 = 12 .
3. 9 · 4 mod 12
solution: or [9 · 4]12 = 12 = 12 .
4. 4 - 7 mod 8
solution: or [4 - 7]8 = [-3]8 = 8 .

6. 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.

7. 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. 8
solution: 8 · 8 = [ 25]8 = 8. Thus, 8 is its own inverse.
2. 10
solution: 10 does not have an inverse, since 5 and 10 have a nontrivial GCD.
3. 7
solution: 7 · 7 = 7 = 7. Thus, 7 is the inverse of 7.
4.  10
solution: 10 does not have an inverse, since 4 and 10 have a nontrivial GCD.

8. 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

9. Use the answer from problem 16 to find multiplicative inverses of:
1. 37
solution: The equation from 16a) gives:
37 = 37·37 - 37·37 .
Since 37 = 37 , we conclude that 37 = 37·37 .
Thus, 37 is the multiplicative inverse of37 .

2. 109
solution: The equation from 16b) gives:
109 = 109·109 - 109·109 .
Since 109 = 109 , we conclude that 109 = [ - 35]109·109 .
Thus, [ - 35]109 is the multiplicative inverse of109 .
We also can write the answer as [ - 35 + 109]109 = 109 Back to the review questions.

For questions: roberts@math.umn.edu