Page 42 - ComputerScience_Class_11

Adding 2E6F and F488, we get 122F7
Removing leftmost 1, we get 22F7.

(22F7) 16

Example 2: (4BC) - (16A5) 16 +1 +1
16
Answer: 15’s complement of 16A5 is E95A. 4 B C
16’s complement of E95A is E95B. + E 9 5 B
E E 1 7
Adding 4BC and E95B, we get EE17.
15’s complement of EE17 is 11E8.
16’s complement of 11E8 is 11E9.

(-11E9) 16

Example 3: (179.E) - (C8.F) 16
16
Answer: 15’s complement of 0C8.F is F37.0 +1 +1
16’s complement of F37.0 is F37.1 1 7 9 . E
+ F 3 7 . 1
Adding 179.E and F37.1, we get 10B0.F
1 0 B 0 . F
Discarding 1 in MSB, we get 0B0.F
(B0.F) 16

Example 4: (6C.29) - (71B.D) 16
16
Answer: 15’s complement of 71B.D0 is 8E4.2F +1 +1
16’s complement is 8E4.30
6 C . 2 9
Adding 6C.29 and 8E4.30, we get 950.59 + 8 E 4 . 3 0
15’s complement of 950.59 is 6AF.A6 9 5 0 . 5 9

16’s complement of 6AF.A6 is 6AF.A7
(-6AF.A7) 16

1.5.3 Hexadecimal Multiplication
It is the method of multiplying hexadecimal numbers. If the product exceeds 15, then the result is divided by 16 to get
the answer. The rightmost digit is written as the answer and the other digit is taken as the carry.

Rules of Hexadecimal Multiplication are:
1. Write the multiplicand and multiplier one below the other, aligning their digits from right to left.
2. Multiply each digit of the multiplicand and multiplier using the rules of hexadecimal multiplication, starting from
the rightmost bit to the leftmost bit. The leftmost bit of each result in the table is the carry. Shift one bit to the left
after each multiplication.
3. Add the results using hexadecimal addition rules.

40 Touchpad Computer Science (Ver. 3.0)-XI