Higher Engineering Mathematics, Sixth Edition

94 Higher Engineering Mathematics

Problem 18. Convert the following decimal

numbers into their hexadecimal equivalents:

(a) 37 10 (b) 108 10

(a)

25

0

Remainder

(^5 5 516)
(^2 5 216)
16 2
16 37
most significant bit least significant bit
Hence 3710 = (^2516)
(b)
6C
0
Remainder
12 5 C 16
(^6 5 616)
16 108
16 6
Hence 10810 =6C 16
Problem 19. Convert the following decimal
numbers into their hexadecimal equivalents:
(a) 162 10 (b) 239 10
(a)
A2
0
Remainder
(^2 5 216)
10 5 A 16
16 162
16 10
Hence 16210 =A2 16
(b)
EF
0
Remainder
15 5 F 16
14 5 E 16
16 239
16 14
Hence 23910 =EF 16
Now try the following exercise
Exercise 42 Further problems on
hexadecimalnumbers
In Problems 1 to 4, convert the given hexadecimal
numbers into their decimal equivalents.

1. E7 16 [231 10 ]2.2C 16 [44 10 ]


2. 98 16 [152 10 ]4.2F1 16 [753 10 ]

In Problems 5 to 8, convert the given decimal

numbers into their hexadecimal equivalents.

5. 54 10 [36 16 ] 6. 200 10 [C8 16 ]


6. 91 10 [5B 16 ] 8. 238 10 [EE 16 ]

(c) Converting from binary to hexadecimal:

The binary bits are arranged in groups of four, start-

ing from right to left, and a hexadecimal symbol is

assigned to each group. For example, the binary num-

ber 1110011110101001is initially grouped in fours as:

(^1110) ︸︷︷︸
E
(^0111) ︸︷︷︸
7
(^1010) ︸︷︷︸
A
(^1001) ︸︷︷︸
9
and a hexadecimal symbol
assigned to each group as above from Table 10.2.
Hence 11100111101010012 =E7A9 16
Problem 20. Convert the following binary
numbers into their hexadecimal equivalents:
(a) 11010110 2 (b) 1100111 2
(a) Grouping bits in fours from the right gives:
(^1101) ︸︷︷︸
D
(^0110) ︸︷︷︸
6
and assigning hexadecimal symbols
to each group gives as above from Table 10.2.
Thus, 110101102 =D6 16
(b) Grouping bits in fours from the right gives:
(^0110) ︸︷︷︸
6
(^0111) ︸︷︷︸
7
and assigning hexadecimal symbols
to each group gives as above from Table 10.2.
Thus, 11001112 = (^6716)
Problem 21. Convert the following binary
numbers into their hexadecimal equivalents:
(a) 11001111 2 (b) 110011110 2
(a) Grouping bits in fours from the right gives:
(^1100) ︸︷︷︸
C
(^1111) ︸︷︷︸
F
and assigning hexadecimal symbols
to each group gives as above from Table 10.2.
Thus, 110011112 =CF 16
(b) Grouping bits in fours from the right gives:
(^0001) ︸︷︷︸
1
(^1001) ︸︷︷︸
9
(^1110) ︸︷︷︸
E
and assigning hexadecimal