Solutions for sixth week’s assignments

Multiplying 26 ⋅ 53 by the Egyptian algorithm:

1 | 53 |

2 | 106 |

4 | 212 |

8 | 424 |

16 | 848 |

Now 26 = 16 + 8 + 2, so 26 ⋅ 53 = 848 + 424 + 106 = \mathbf{1378}.

Computing 33 ⋅ 81:

1 | 81 |

2 | 162 |

4 | 324 |

8 | 648 |

16 | 1296 |

32 | 2592 |

Because 33 = 32 + 1, 33 ⋅ 81 = 2592 + 81 = \mathbf{2673}.

- Problem 2
- 925 = 9 ⋅ 1{0}^{2} + 2 ⋅ 1{0}^{1} + 5 ⋅ 1{0}^{0}
- Problem 3
- 3774 = 3 ⋅ 1{0}^{3} + 7 ⋅ 1{0}^{2} + 7 ⋅ 1{0}^{1} + 4 ⋅ 1{0}^{0}
- Problem 5
- 4 ⋅ 1{0}^{3} + 9 ⋅ 1{0}^{2} + 2 ⋅ 1{0}^{1} + 4 ⋅ 1{0}^{0}
- Problem 6
- 5 ⋅ 1{0}^{4} + 2 ⋅ 1{0}^{3} + 1 ⋅ 1{0}^{2} + 1 ⋅ 1{0}^{1} + 8 ⋅ 1{0}^{0}
- Problem 11
- 6209
- Problem 12
- 503568

- Problem 2
- 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 13, 14, 15, 16, 17, 20, 21, 22, 23, 24
- Problem 7
- \mathbf{\text{B}6{\text{E}}_{16}}, \text{B}6{\text{F}}_{16}, \mathbf{\text{B}7{0}_{16}}
- Problem 8
- \mathbf{1011{0}_{2}}, 1011{1}_{2}, \mathbf{1100{0}_{2}}
- Problem 19
- 3{\text{B}\text{C}}_{16} = (3 ⋅ 16 + 11) ⋅ 16 + 12 = 956
- Problem 20
- 3443{2}_{5} = (((3\dot{5} + 4) ⋅ 5 + 4) ⋅ 5 + 3) ⋅ 5 + 2 = 2492
- Problem 21
- 236{6}_{7} = ((2 ⋅ 7 + 3) ⋅ 7 + 6) ⋅ 7 + 6 = 881
- Problem 22
- 10110111{0}_{2} = (((((((1⋅2+0)⋅2+1)⋅2+1)⋅2+0)⋅2+1)⋅2+1)⋅2+1)⋅2+0 = 366
- Problem 37
- 586 = 512 + 64 + 8 + 2 = {2}^{9} + {2}^{6} + {2}^{3} + {2}^{1} = 100100101{0}_{2}
- Problem 38
- 12888 = 3 ⋅ 4096 + 1 ⋅ 512 + 1 ⋅ 64 + 3 ⋅ 8 = 3 ⋅ {8}^{4} + 1 ⋅ {8}^{3} + 1 ⋅ {8}^{2} + 3 ⋅ 8 = 3113{0}_{8}
- Problem 39
- 8407 = 10211210{1}_{3}
- Problem 40
- 11028 = 223011{0}_{4}
- Problem 57
- 9 ⋅ 1{2}^{2} + 10 ⋅ 12 + 11 = 1427

Take a familiar incomplete integer, -679-, and express it as a sum of the digits times powers of ten using variables {x}_{0} and {x}_{4} for the digits in the blanks. Simplify to the form of {x}_{4} ⋅ 1{0}^{4} + {x}_{0} ⋅ 1{0}^{0} + z, where z is a single number in positional form (a sequence of digits). Does 72 divide z? Does 8 divide z? Does 9 divide z? Remember that 72 = 8 ⋅ 9. We will use this example again in the next chapter.

We can expand -679- to be {x}_{4} ⋅ 1{0}^{4} + 6 ⋅ 1{0}^{3} + 7 ⋅ 1{0}^{2} + 9 ⋅ 10 + {x}_{0} = {x}_{4} ⋅ 1{0}^{4} + {x}_{0} + 6790, so \mathbf{z = 6790}. Unfortunately, none of the numbers provided divide cleanly into 6790. Jumping to the next chapter, 6790 = 94 ⋅ 72 + 22 = 848 ⋅ 8 + 6 = 754 ⋅ 9 + 4.

Multiplication:

- 47 ⋅ 3 = (4 ⋅ 10 + 7) ⋅ 3 = 12 ⋅ 10 + 21 = 141.

Or in table form:4 7 ⋅\quad 3 2 1 1 2 1 4 1 - 47 ⋅ 13 = (4 ⋅ 10 + 7) ⋅ (10 + 3) = (4 ⋅ 1{0}^{2} + 7 ⋅ 10) + (12 ⋅ 10 + 21) = 470 + 141 = 611.

Or in table form:4 7 ⋅\quad 1 3 2 1 1 2 1 4 1 4 7 6 1 1 - 47 ⋅ 23 = (4 ⋅ 10 + 7) ⋅ (2 ⋅ 10 + 3) = (8 ⋅ 1{0}^{2} + 14 ⋅ 10) + (12 ⋅ 10 + 21) = 940 + 141 = 1081.

Or in table form:4 7 ⋅\quad 2 3 2 1 1 2 1 4 1 1 4 2 4 0 8 1 0 8 1

Addition:

- 47 + 52 = (4 + 5) ⋅ 10 + (7 + 2) = 9 ⋅ 10 + 9 = 99.
- 47+53 = (4+5)⋅10+(7+3) = \mathbf{9 ⋅ 10 + 10} = \mathbf{10 ⋅ 10 + 0} = 1⋅1{0}^{2}+0⋅10+0 = 100. Both bold forms are redundant intermediate representations.
- 47+54 = (4+5)⋅10+(7+4) = \mathbf{9 ⋅ 10 + 11} = \mathbf{10 ⋅ 10 + 1} = 1⋅1{0}^{2}+0⋅10+1 = 101. Both bold forms are redundant intermediate representations.

Subtraction:

- 7 - 19 = (0 - 1) ⋅ 10 + (7 - 9) = \mathbf{-1 ⋅ 10 + -2⋅} = -1 ⋅ (1 ⋅ 10 + 2) = -12. The bold form is a redundant intermediate representation.
- 19 - 19 = (1 - 1) ⋅ 10 + (9 - 9) = 0 + 0 = 0.
- 20 - 19 = (2 - 1) ⋅ 10 + (0 - 9) = \mathbf{1 ⋅ 10 + -9} = 0 ⋅ 10 + (10 - 9) = 1. The bold form is a redundant intermediate representation.
- 29 - 19 = (2 - 1) ⋅ 10 + (9 - 9) = 1 ⋅ 10 + 0 = 10.