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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:
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 .