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.

- Find prime factorizations of the following numbers:
- 28

2*solution:*^{2}·7 - 68

2*solution:*^{2}·17 - 323

17·19*solution:* - 108

2*solution:*^{2}·3^{3} - 109

it is prime. Check that numbers*solution:*__<__11 don't divide it.

- Using the prime factorizations, find the following greatest common divisors and least common multiples:
- GCD(28,68) and LCM(28,68)

GCD = 2*solution:*^{2}and LCM = 2^{2}·7·17 - GCD(28,108) and LCM(28,108)

GCD = 2*solution:*^{2}and LCM = 2^{2}·3^{3}·7 - GCD(28,323)

GCD = 1*solution:* - GCD(68,323)

GCD = 17*solution:* - GCD(28,109)

GCD = 1*solution:*

- Use the Euclidean algorithm to find the following GCD's:
- GCD(16,37)

37 = 2·16 + 5, so that GCD(16,37) = GCD(16,5)*solution:*

16 = 3·5 + 1, so that GCD(16,5) = GCD(5,1) = 1.

We conclude that GCD(16,37) = 1 - GCD(28,109)

109 = 3·28 + 25, so that GCD(28,109) = GCD(28,25)*solution:*

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 - GCD(1003,1999)

1999 = 1003 + 996, so that GCD(1003,1999) = GCD(1003,996)*solution:*

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

- In each case, find the next five numbers in the given base:
- 11011
_{two}

the next 5 numbers are 11100*solution:*_{two}, 11101_{two}, 11110_{two}, 11111_{two}, 100000_{two}. - 2221
_{three}

the next 5 numbers are 2222*solution:*_{three}, 10000_{three}, 10001_{three}, t0002_{three}, 10010_{three}. - 2AD
_{ sixteen}

the next 5 numbers are 2AE*solution:*_{sixteen}, 2AF_{sixteen}, 2B0_{sixteen}, 2B1_{sixteen}, 2B2_{sixteen}.

- Convert to bases 2, 3, and 16:
- 79

79 = 64 + 8 + 4 + 2 + 1 = 2*solution:*^{6}+ 2^{3}+ 2^{2}+ 2^{1}+ 2^{0}= 1001111_{two}

79 = 2·27 + 2·9 + 2·3 + 1 = 2·3^{3}+ 2·3^{2}+ 2·3^{1}+ 1·3^{0}= 2221_{three}

79 = 4·16 + 15 = 4F_{sixteen}

. - 257

257 = 256 + 1 = 2*solution:*^{8}+ 2^{0}= 100000001_{two}

257 = 243 + 9 + 3 + 2 = 3^{5}+ 3^{2}+ 3^{1}+ 2·3^{0}= 100112_{three}

257 = 256 + 1 = 16^{2}+ 1 = 101_{sixteen}.

- Convert to base 10:
- 100011
_{two}

100011*solution:*_{two}= 2^{5}+ 2^{1}+ 2^{0}= 32 + 2 + 1 = 35 - 177
_{nine}

177*solution:*_{nine}= 9^{2}+ 7·9 + 7 = 81 + 63 + 7 = 151 - BBB
_{ twelve}

BBB*solution:*_{ twelve}= 11·12^{2}+ 11·12 + 11 = 1584 + 132 + 11 = 1727

12*check:*^{3}- 1 = 1728 - 1 = 1727

- Do the indicated conversions:
- 110111011
_{two}to base eight

(110)(111)(011)*solution:*_{two}= 673_{eight} - 21201
_{three}to base nine

(02)(12)(01)*solution:*_{three}= 251_{nine}.

- 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:
- Add in base eight: 6142
_{eight}and 2457_{eight}eight

*solution:*

- Subtract in base eight: 651
_{eight}from 6142_{eight}

*solution:*

- Multiply in base five: 423
_{five}by 214_{ five}

*solution:*

Back to the review questions.

For questions: `roberts@math.umn.edu
`