Download Number Systems – Conversion & Math Practice Problems and more Lecture notes Number Theory in PDF only on Docsity!
Conversion Problems
- Convert each of the following binary numbers to octal, decimal, and hexadecimal formats.
(111011101) 2
(10101010111) 2
(111100000) 2
- Convert each of the following octal numbers to binary, decimal, and hexadecimal formats.
(3754) 8
(7777) 8
(247) 8
- Convert each of the following decimal numbers to binary, octal, and hexadecimal formats.
(3479) 10
(642) 10
(555) 10
- Convert each of the following hexadecimal numbers to binary, octal, and decimal formats.
(4FB2) 16
(88BAE) 16
(DC4) 16
Number Systems – Conversion & Math Practice Problems
Math Problems
- Perform each of the addition operations indicated below.
(1001011) 2 + (11101) 2
(4556) 8 + (1245) 8
(BCD) 16 + (A34) 16
- Form the two’s complement of each of the following binary numbers.
(111011101110) 2
(11111111000100) 2
(100000000) 2
(1010101010111) 2
- Perform each of the subtraction operations indicated below using addition and the two’s complement of the subtrahend.
(100101) 2 – (11011) 2
(1101011) 2 – (111010) 2
(1110111) 2 – (10110111) 2
to decimal: =(7x8^3 ) + (7x8^2 ) + (7x8^1 ) + (7x8^0 )
7. Convert each of the following decimal numbers to binary, octal, and
- (7777)
- to binary: = (111 111 111 111) - = 3584 + 448 + 56 + - = (4095)
- to hexadecimal: = (1111 1111 1111) 2 = (FFF)
- (247)
- to binary: = (10 100 111) - = 128 + 32 + to decimal: =(2x8^2 ) + (4x8^1 ) + (7x8^0 ) - = (167)
- to hexadecimal: = (1010 0111) 2 = (A7)
- (3479) hexadecimal formats.
- to binary: = 3479 ÷ 2 = 1739 rem = - 1739 ÷ 2 = 869 rem = - 869 ÷ 2 = 434 rem = - 434 ÷ 2 = 217 rem = - 217 ÷ 2 = 108 rem = - 108 ÷ 2 = 54 rem = - 54 ÷ 2 = 27 rem = - 27 ÷ 2 = 13 rem = - 13 ÷ 2 = 6 rem = - 6 ÷ 2 = 3 rem = - 3 ÷ 2 = 1 rem = - 1 ÷ 2 = 0 rem =
- reading bottom to top of remainders = (110110010111)
- to octal: = 3479 ÷ 8 = 434 rem = - 434 ÷ 8 = 54 rem = - 54 ÷ 8 = 6 rem = - 6 ÷ 8 = 0 rem =
- reading bottom to top of remainders = (6627)
- to hexadecimal: = 3479 ÷ 16 = 217 rem = - 217 ÷ 16 = 13 rem =
- reading bottom to top of remainders = (D97) 13 ÷ 16 = 0 rem = 13 (D)
- (642)
- to binary: = 642 ÷ 2 = 321 rem = - 321 ÷ 2 = 160 rem = - 160 ÷ 2 = 80 rem = - 80 ÷ 2 = 40 rem = - 40 ÷ 2 = 20 rem = - 20 ÷ 2 = 10 rem = - 10 ÷ 2 = 5 rem = - 5 ÷ 2 = 2 rem = - 2 ÷ 2 = 1 rem = - 1 ÷ 2 = 0 rem =
- reading bottom to top of remainders = (1010000010)
- to octal: = 642 ÷ 8 = 80 rem = - 80 ÷ 8 = 10 rem = - 10 ÷ 8 = 1 rem = - 1 ÷ 8 = 0 rem =
- reading bottom to top of remainders = (1202)
- to hexadecimal: = 642 ÷ 16 = 40 rem = - 40 ÷ 16 = 2 rem = - 2 ÷ 16 = 0 rem =
- reading bottom to top of remainders = (282)
- (555)
- to binary: = 555 ÷ 2 = 277 rem = - 277 ÷ 2 = 138 rem = - 138 ÷ 2 = 69 rem = - 69 ÷ 2 = 34 rem = - 34 ÷ 2 = 17 rem = - 17 ÷ 2 = 8 rem = - 8 ÷ 2 = 4 rem = - 4 ÷ 2 = 2 rem = - 2 ÷ 2 = 1 rem =
Math Problems
4. Perform each of the addition operations indicated below.
(1001011) 2 + (11101) 2
11111 carry 1001011
1101000
111 carry 4556
(BCD) 16 + (A34) 16
11 carry BCD
5. Form the two’s complement of each of the following binary numbers.
(111011101110) 2
technique #1: form one’s complement and add 1 000100010001
technique #2: leave least significant 0’s unchanged up to an including first least significant 1 – then complement all remaining bits.
000100010010
two’s complement is: 00000000111100
(100000000) 2
twos’ complement is: 100000000 (note that it is the same!)
two’s complement is: 0101010101001
6. Perform each of the subtraction operations indicated below using addition and the two’s complement of the subtrahend.
(100101) 2 – (11011) 2
1 1 1 carry 100101
1001010 carry out of MSB is ignored result is (1010) 2 = (10) 10
(1101011) 2 – (111010) 2
111 carry 1101011
10110001 carry out of MSB is ignored result is (110001) 2 = (49) 10
1111111 carry 01110111
- 01001001 11000000 no carry out of MSB – result is in 2’s comp and is negative, result is (1000000) 2 = (-64) 10
