Computations in Base five & six
Homework
Answers
1.
Add the following
numbers in base five.
1
a.
102
+ 34 1 0 2
+ 3 4
1 4 1
1 1 1 1 1
3 2 4 0 4
b.
32404
+ 23043 + 2 3
0 4 3
1 1 1 0 0 2
2.
Add
the following numbers in base six.
1 1
2 5
4
a.
254
+ 43 +
4 3
3 4 1
1 1 1 1 1
4 5
0 2 3
b. 45023 +
23553 + 2
3 5 5 3
1 1 3
0 2 0
3.
Subtract
the following numbers in base five. Do the problem twice – once by
borrowing and once by equal addition.
1 2 3
1 2 3
(0) (12) 3
a.
123
– 42 -
4 2 -
4 2 - 4
2
3 1
1 2
3 1 (12)
3
-
4 2 -(1) 4 2
3 1
4 0 2 (3) (4) (12) (3) (4) (12)
b.
402
– 133 - 1
3 3 -
1 3 3 - 1
3 3
2 1 4
4 0 2
4 (10) (12)
4 (10) (12)
-
1 3 3 - (2)
(4) 3 - (2) (4)
3
2 1 4
4 1
2 3 (3) (10)(12) 3
(3) (10)(12) 3
c.
4123
– 1233 -
1 2 3 3 - 1
2 3 3 -
1 2 3
3
2 3
4 0
4
1 2 3
4 (11) (12) 3 4 (11) (12) 3
- 1
2 3 3 - (2) (3) 3 3 - (2)
(3) 3 3
2 3
4 0
4.
Subtract
the following numbers in base six. Do the problem twice – once by
borrowing and once by equal addition.
a.
352
– 45 3
5 2 3
(4) (12) 3 (4) (12)
-
4 5 -
4 5 -
4 5
3 0 3
3 5
2 3 5 (12) 3
5 (12)
- 4
5 -
(5) 5 -
(5) 5
3 0 3
4 0 2
1 (3) (5)(11)(11) (3) (5)(11)(11)
b.
4021
– 3153 - 3
1 5 3
- 3 1
5 3 -
3 1 5 3
4 2 4
4 0
2 1 4 (10)
2 (11) 4
(10) 2 (11)
- 3
1 5 3 -
(4) (2) (0) 3 - (4)
(2) (0) 3
4 2
4
1 5 3 4 2 1(4)(12)(13)(12) 1(4)(12)(13)(12)
c.
15342
– 4553 - 4
5 5 3 - 4
5 5 3 -
4 5 5 3
1 0 3
4 5
1 5
3 4 2
1 5 3
4 2 1
5 3 4 (12)
- 4
5 5 3 - (5) (0)
(0) 3 -
(5) (0) (0) 3
1 0 3
4 5
5.
Multiply
the following numbers in base five. Do the problem twice – once by the
standard algorithm and once by the lattice method.
a.
21
x 43
2 1
x 4 3
1 1 3
1 3 4 .
2 0
0 3
b.
3413
x 23
3 4
1 3
x 2 3
2 1 2
4 4
1 2 3 3 1 .
2 0 0 1
0 4
c.
1234 x 432
1 2
3 4
x 4 3 2
3 0 2 3
4 3 1 2
1 1 1 0
1 .
1
2
1 1 2
4 3
6.
Multiply
the following numbers in base six. Do the problem twice – once by the
standard algorithm and once by the lattice method.
2
5
a.
25
x 43 x 4 3
1 2 3
2 3 2
2 0
4 3
1 3 5
2
x 5
3
b.
1352
x 53 4
5 4 0
1
2 1 2
4
1 3
0 2 2 0
c.
3524
x 452
3 5 2 4
x 4 5 2
1 1 4
5 2
2 1 3
1 2
2 3 3 4 4 .
3 1
0 3 4
1 2
7.
Compute
the partial product table for 143 in both
base five and base six.
1 4 3 x 1
= 1 4 3 1 4
3 x 1
= 1 4 3
1 4 3 x
2 = 3 4 1 1
4 3 x
2 = 3
3 0
1 4 3 x
3 = 1 0 3 4 1 4
3 x 3
= 5 1 3
1 4 3 x
4 = 1 2 3
2 1 4
3 x 4
= 1 1 0 0
Base five 1 4 3
x 5 =
1 2 4 3
Base six
8. Use these tables to compute the following divisions.
a. 
base five
. 1 2
1 4 3 )
2 3 4 1
1 4
3
4 1 1
3 4
1
2 0
b. 
base five . 1 1 4
1 4 3 ) 2
3 1 4 1
1 4 3
3 3 4
1 4 3
1 4 1 1
1 2 3
2
1 2
4
c.
base six . 1 4
1 4 3
) 2 5
4 1
1 4 3
1 1 1 1
1 1
0 0
1 1
d.
base
six . 1 4
0 0
1 4 3
) 2 5
3 1 4 1
1 4 3
1 1 0 1
1 1
0 0
1 4 1
e. 
base six . 1 3
5
1 4 3
) 2 5
2 4 1
1 4 3
1 0 5 4
5 1 3
1 4 1 1
1 2 4
3
1 2 4