 ### Solutions to chapter 5 and 6 review questions

#### (Problems 1 through 8) NOTE: It is also strongly recommended to review past homework problems and your notes from group work in class. The problems on this page do not represent every type of question that could occur on the test.

1. Find prime factorizations of the following numbers:
1. 28
solution: 22·7
2. 68
solution: 22·17
3. 323
solution: 17·19
4. 108
solution: 22·33
5. 109
solution: it is prime. Check that numbers < 11 don't divide it.

2. Using the prime factorizations, find the following greatest common divisors and least common multiples:
1. GCD(28,68) and LCM(28,68)
solution: GCD = 22 and LCM = 22·7·17
2. GCD(28,108) and LCM(28,108)
solution: GCD = 22 and LCM = 22·33·7
3. GCD(28,323)
solution: GCD = 1
4. GCD(68,323)
solution: GCD = 17
5. GCD(28,109)
solution: GCD = 1

3. Use the Euclidean algorithm to find the following GCD's:
1. GCD(16,37)
solution: 37 = 2·16 + 5, so that GCD(16,37) = GCD(16,5)
16 = 3·5 + 1, so that GCD(16,5) = GCD(5,1) = 1.
We conclude that GCD(16,37) = 1
2. GCD(28,109)
solution: 109 = 3·28 + 25, so that GCD(28,109) = GCD(28,25)
28 = 25 + 3, so that GCD(28,25) = GCD(25, 3)
25 = 8·3 + 1, so that GCD(25, 3) = GCD(3,1) = 1.
We conclude that GCD(28,109) = 1
3. GCD(1003,1999)
solution: 1999 = 1003 + 996, so that GCD(1003,1999) = GCD(1003,996)
1003 = 996 + 7 so that GCD(1003,996) = GCD(996, 7)
996 = 142·7 + 3, so that GCD(996, 7) = GCD(7, 3)
7 = 3·2 + 1, so that GCD(7,3) = GCD(3,1) = 1
We conclude that GCD(1003,1999) = 1

4. In each case, find the next five numbers in the given base:
1. 11011two
solution: the next 5 numbers are 11100two, 11101two, 11110two, 11111two, 100000two .
2. 2221three
solution: the next 5 numbers are 2222three, 10000three, 10001three, t0002three, 10010three .
solution: the next 5 numbers are 2AEsixteen, 2AFsixteen, 2B0sixteen, 2B1sixteen, 2B2sixteen .

5. Convert to bases 2, 3, and 16:
1. 79
solution: 79 = 64 + 8 + 4 + 2 + 1 = 26 + 23 + 22 + 21 + 20 = 1001111two

79 = 2·27 + 2·9 + 2·3 + 1 = 2·33 + 2·32 + 2·31 + 1·30 = 2221three

79 = 4·16 + 15 = 4Fsixteen
.
2. 257
solution: 257 = 256 + 1 = 28 + 20 = 100000001two

257 = 243 + 9 + 3 + 2 = 35 + 32 + 31 + 2·30 = 100112three

257 = 256 + 1 = 162 + 1 = 101sixteen .

6. Convert to base 10:
1. 100011two
solution: 100011two = 25 + 21 + 20 = 32 + 2 + 1 = 35
2. 177nine
solution: 177nine = 92 + 7·9 + 7 = 81 + 63 + 7 = 151
3. BBB twelve
solution: BBB twelve = 11·122 + 11·12 + 11 = 1584 + 132 + 11 = 1727
check: 123 - 1 = 1728 - 1 = 1727

7. Do the indicated conversions:
1. 110111011two to base eight
solution: (110)(111)(011)two = 673eight
2. 21201three to base nine
solution: (02)(12)(01)three = 251nine .

8. Do the following calculations in the indicated bases. Show your work, including which numbers were carried. Check your answer by converting to base 10 and doing appropriate base 10 calculations:
1. Add in base eight: 6142eight and 2457eight eight

solution: 2. Subtract in base eight: 651eight from 6142eight

solution: 3. Multiply in base five: 423five by 214 five

solution:  Back to the review questions.

For questions: roberts@math.umn.edu