1.11 General Number Bases Solutions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
B
A
A
B
B
A
B
A
B
B
A
B
B
B
A
B
A
A
Explanations
1.
2.
3.
4.
5.
6.
Each digit in a base B is weighted by a power of B.
The values are 0 to 5-1. A 5 in a digit would instead be a 1 in the next higher digit.
1×52 + 0×51 + 4×50 = 25 + 4 = 29
Each digit can be 0 to 8-1, or 0 to 7. The rightmost digit should be 7, with a 1 carried to the next
digit. So the number should be 137.
7.
8.
9.
10.
11.
1×52 + 1×51 + 1×50 = 25 + 5 + 1 = 31
2×302 + 0×301 + 0×300 = 2 × 900 = 1800
29 / 3 = 9 rem 2. Thus rightmost digit is 2.
29 / 3 = 9 rem 2.
9 / 3 = 3 rem 0.
3 / 3 = 1 rem 0.
1 / 3 = 0 rem 1.
Concatenating the remainders in reverse yields: 1002.
12. Converting 200 base 3 into decimal gives 18
converting 18 base 10 into base 9 gives 20
13.
14.
15. 23 / 7 = 3 remainder 2
3 / 7 = 0 remainder 3
Concatenating the remainders in reverse yields: 32
16. 27 / 12 = 2 remainder 3
2 / 12 = 0 remainder 2
Concatenating the remainders in reverse yields: 23
17. Converting 32 base 4 into decimal we get 14,
Now convert to base 5:
14 / 5 = 2 remainder 4
2 / 5 = 0 remainder 2
Concatenating the remainders in reverse yields: 24
18. Converting 34 base 6 to base 10 we get 22
Now convert to base 5:
22 / 5 = 4 remainder 2
4 / 5 = 0 remainder 4
Concatenating the remainders in reverse yields: 42