Exercise DET-01.02 – Converting numbers between number systems (Solution)

Prof. Dr. Thorsten Jungmann (last updated 2026-11-02)

Relates to DET-01 Introduction to the module and its content

Task

Convert each of the following numbers into the stated number system.

Binary → decimal: a) $1110010 1_{2}$ , b) $10110100110 1_{2}$

Octal → decimal: c) $25 3_{8}$ , d) $472 6_{8}$

Hexadecimal → decimal: e) $9E_{16}$ , f) $C3A_{16}$

Decimal → binary: g) $18 7_{10}$ , h) $285 4_{10}$

Decimal → octal: i) $20 1_{10}$ , j) $389 0_{10}$

Decimal → hexadecimal: k) $22 2_{10}$ , l) $278 8_{10}$

Binary → octal: m) $1001110 1_{2}$ , n) $11101011000 1_{2}$

Octal → binary: o) $36 5_{8}$ , p) $304 2_{8}$

Binary → hexadecimal: q) $1010011 1_{2}$ , r) $11010011100 0_{2}$

Hexadecimal → binary: s) $5B_{16}$ , t) $8F2_{16}$

Octal → hexadecimal: u) $31 5_{8}$ , v) $652 4_{8}$

Hexadecimal → octal: w) $E9_{16}$ , x) $A7E_{16}$

◀️ Back to the task

Solution

Given

Given explicitly:

Binary → decimal: $1110010 1_{2}$ , $10110100110 1_{2}$
Octal → decimal: $25 3_{8}$ , $472 6_{8}$
Hexadecimal → decimal: $9E_{16}$ , $C3A_{16}$
Decimal → binary: $18 7_{10}$ , $285 4_{10}$
Decimal → octal: $20 1_{10}$ , $389 0_{10}$
Decimal → hexadecimal: $22 2_{10}$ , $278 8_{10}$
Binary → octal: $1001110 1_{2}$ , $11101011000 1_{2}$
Octal → binary: $36 5_{8}$ , $304 2_{8}$
Binary → hexadecimal: $1010011 1_{2}$ , $11010011100 0_{2}$
Hexadecimal → binary: $5B_{16}$ , $8F2_{16}$
Octal → hexadecimal: $31 5_{8}$ , $652 4_{8}$
Hexadecimal → octal: $E9_{16}$ , $A7E_{16}$

Known:

Into the decimal number system (DET-01, 7.1 to 7.3): multiply each digit by its weight and add up the products
From the decimal number system (DET-01, 7.4 to 7.6): repeated division by the base of the target system; the remainders, read from bottom to top, form the result
Between binary, octal and hexadecimal (DET-01, 7.7): one octal digit stands for three bits, one hexadecimal digit for four bits; group the bits from the right; between octal and hexadecimal, the way leads via binary
Weights: $2^{0} = 1$ to $2^{11} = 2048$ ; $8^{0} = 1$ , $8^{1} = 8$ , $8^{2} = 64$ , $8^{3} = 512$ ; $1 6^{0} = 1$ , $1 6^{1} = 16$ , $1 6^{2} = 256$
Values of the hexadecimal digits: A = 10, B = 11, C = 12, D = 13, E = 14, F = 15

Required

a) to x) each number in the stated number system

Binary → decimal

Multiply each digit by its weight and add up the products. In binary, a 0 contributes nothing and a 1 contributes its weight.

a) $1110010 1_{2}$

1110010 1_{2} = 1 \cdot 2^{7} + 1 \cdot 2^{6} + 1 \cdot 2^{5} + 0 \cdot 2^{4} + 0 \cdot 2^{3} + 1 \cdot 2^{2} + 0 \cdot 2^{1} + 1 \cdot 2^{0} = 128 + 64 + 32 + 4 + 1 = 22 9_{10}

Check via hexadecimal: in groups of four bits, $1110010 1_{2} = E5_{16}$ , and $E5_{16} = 14 \cdot 16 + 5 = 229$ ✓

b) $10110100110 1_{2}$

10110100110 1_{2} = 1 \cdot 2^{11} + 0 \cdot 2^{10} + 1 \cdot 2^{9} + 1 \cdot 2^{8} + 0 \cdot 2^{7} + 1 \cdot 2^{6} + 0 \cdot 2^{5} + 0 \cdot 2^{4} + 1 \cdot 2^{3} + 1 \cdot 2^{2} + 0 \cdot 2^{1} + 1 \cdot 2^{0} = 2048 + 512 + 256 + 64 + 8 + 4 + 1 = 289 3_{10}

Check via hexadecimal: in groups of four bits, $10110100110 1_{2} = B4D_{16}$ , and $B4D_{16} = 11 \cdot 256 + 4 \cdot 16 + 13 = 2893$ ✓

Octal → decimal

Multiply each digit by its weight and add up the products.

c) $25 3_{8}$

25 3_{8} = 2 \cdot 8^{2} + 5 \cdot 8^{1} + 3 \cdot 8^{0} = 128 + 40 + 3 = 17 1_{10}

Check via binary and hexadecimal: $25 3_{8} = 1010101 1_{2} = AB_{16}$ , and $AB_{16} = 10 \cdot 16 + 11 = 171$ ✓

d) $472 6_{8}$

472 6_{8} = 4 \cdot 8^{3} + 7 \cdot 8^{2} + 2 \cdot 8^{1} + 6 \cdot 8^{0} = 2048 + 448 + 16 + 6 = 251 8_{10}

Check via binary and hexadecimal: $472 6_{8} = 10011101011 0_{2} = 9D6_{16}$ , and $9D6_{16} = 9 \cdot 256 + 13 \cdot 16 + 6 = 2518$ ✓

Hexadecimal → decimal

Multiply each digit by its weight and add up the products. The letters count with their values.

e) $9E_{16}$

9E_{16} = 9 \cdot 1 6^{1} + 14 \cdot 1 6^{0} = 144 + 14 = 15 8_{10}

Check by repeated division by 16:

Division	Quotient	Remainder
$158/16$	$9$	14 = E
$9/16$	$0$	9

Read from bottom to top, the remainders give back $9E_{16}$ ✓

f) $C3A_{16}$

C3A_{16} = 12 \cdot 1 6^{2} + 3 \cdot 1 6^{1} + 10 \cdot 1 6^{0} = 3072 + 48 + 10 = 313 0_{10}

Check by repeated division by 16:

Division	Quotient	Remainder
$3 130/16$	$195$	10 = A
$195/16$	$12$	3
$12/16$	$0$	12 = C

Read from bottom to top, the remainders give back $C3A_{16}$ ✓

Decimal → binary

Divide the number repeatedly by 2 and note the remainders until the quotient is 0. The first remainder is the least significant digit; read from bottom to top, the remainders form the result.

g) $18 7_{10}$

Division	Quotient	Remainder
$187/2$	$93$	1 (LSB)
$93/2$	$46$	1
$46/2$	$23$	0
$23/2$	$11$	1
$11/2$	$5$	1
$5/2$	$2$	1
$2/2$	$1$	0
$1/2$	$0$	1 (MSB)

Read from bottom to top: $18 7_{10} = 1011101 1_{2}$

Check: $1011101 1_{2} = 128 + 32 + 16 + 8 + 2 + 1 = 187$ ✓

h) $285 4_{10}$

Division	Quotient	Remainder
$2 854/2$	$1427$	0 (LSB)
$1 427/2$	$713$	1
$713/2$	$356$	1
$356/2$	$178$	0
$178/2$	$89$	0
$89/2$	$44$	1
$44/2$	$22$	0
$22/2$	$11$	0
$11/2$	$5$	1
$5/2$	$2$	1
$2/2$	$1$	0
$1/2$	$0$	1 (MSB)

Read from bottom to top: $285 4_{10} = 10110010011 0_{2}$

Check: $10110010011 0_{2} = 2048 + 512 + 256 + 32 + 4 + 2 = 2854$ ✓

Alternatively: subtracting powers of two

Subtract the largest power of two that fits and write a 1 at its position; repeat with the rest. Here, $2854 = 2048 + 512 + 256 + 32 + 4 + 2$ , i.e. 1s at the positions 11, 9, 8, 5, 2, 1 and 0s at all others.

Decimal → octal

Divide the number repeatedly by 8 and note the remainders until the quotient is 0. The first remainder is the least significant digit; read from bottom to top, the remainders form the result.

i) $20 1_{10}$

Division	Quotient	Remainder
$201/8$	$25$	1
$25/8$	$3$	1
$3/8$	$0$	3

Read from bottom to top: $20 1_{10} = 31 1_{8}$

Check: $31 1_{8} = 3 \cdot 64 + 8 + 1 = 201$ ✓

j) $389 0_{10}$

Division	Quotient	Remainder
$3 890/8$	$486$	2
$486/8$	$60$	6
$60/8$	$7$	4
$7/8$	$0$	7

Read from bottom to top: $389 0_{10} = 746 2_{8}$

Check: $746 2_{8} = 7 \cdot 512 + 4 \cdot 64 + 6 \cdot 8 + 2 = 3890$ ✓

Decimal → hexadecimal

Divide the number repeatedly by 16 and note the remainders until the quotient is 0. The first remainder is the least significant digit; read from bottom to top, the remainders form the result. Remainders from 10 to 15 are written as A to F.

k) $22 2_{10}$

Division	Quotient	Remainder
$222/16$	$13$	14 = E
$13/16$	$0$	13 = D

Read from bottom to top: $22 2_{10} = DE_{16}$

Check: $DE_{16} = 13 \cdot 16 + 14 = 222$ ✓

l) $278 8_{10}$

Division	Quotient	Remainder
$2 788/16$	$174$	4
$174/16$	$10$	14 = E
$10/16$	$0$	10 = A

Read from bottom to top: $278 8_{10} = AE4_{16}$

Check: $AE4_{16} = 10 \cdot 256 + 14 \cdot 16 + 4 = 2788$ ✓

Binary → octal

Group the bits from the right in groups of three, add leading zeros to fill the leftmost group, and replace each group by its octal digit.

m) $1001110 1_{2}$

Groups of three bits	010	011	101
Octal digits	2	3	5

$1001110 1_{2} = 23 5_{8}$ (with one leading zero in the leftmost group)

Check via the decimal value: $1001110 1_{2} = 128 + 16 + 8 + 4 + 1 = 157$ and $23 5_{8} = 2 \cdot 64 + 3 \cdot 8 + 5 = 157$ ✓

n) $11101011000 1_{2}$

Groups of three bits	111	010	110	001
Octal digits	7	2	6	1

$11101011000 1_{2} = 726 1_{8}$

Check via the decimal value: $11101011000 1_{2} = 2048 + 1024 + 512 + 128 + 32 + 16 + 1 = 3761$ and $726 1_{8} = 7 \cdot 512 + 2 \cdot 64 + 6 \cdot 8 + 1 = 3761$ ✓

Octal → binary

Replace each octal digit by its group of three bits.

o) $36 5_{8}$

Octal digits	3	6	5
Groups of three bits	011	110	101

Without the leading zero and grouped in nibbles for readability: $36 5_{8} = 1111010 1_{2}$

Check via the decimal value: $36 5_{8} = 3 \cdot 64 + 6 \cdot 8 + 5 = 245$ and $1111010 1_{2} = 128 + 64 + 32 + 16 + 4 + 1 = 245$ ✓

p) $304 2_{8}$

Octal digits	3	0	4	2
Groups of three bits	011	000	100	010

Without the leading zero and grouped in nibbles for readability: $304 2_{8} = 1100010001 0_{2}$

Check via the decimal value: $304 2_{8} = 3 \cdot 512 + 4 \cdot 8 + 2 = 1570$ and $1100010001 0_{2} = 1024 + 512 + 32 + 2 = 1570$ ✓

Binary → hexadecimal

Group the bits from the right in groups of four, add leading zeros to fill the leftmost group, and replace each group by its hexadecimal digit.

q) $1010011 1_{2}$

Groups of four bits	1010	0111
Hexadecimal digits	A	7

$1010011 1_{2} = A7_{16}$

Check via the decimal value: $1010011 1_{2} = 128 + 32 + 4 + 2 + 1 = 167$ and $A7_{16} = 10 \cdot 16 + 7 = 167$ ✓

r) $11010011100 0_{2}$

Groups of four bits	1101	0011	1000
Hexadecimal digits	D	3	8

$11010011100 0_{2} = D38_{16}$

Check via the decimal value: $11010011100 0_{2} = 2048 + 1024 + 256 + 32 + 16 + 8 = 3384$ and $D38_{16} = 13 \cdot 256 + 3 \cdot 16 + 8 = 3384$ ✓

Hexadecimal → binary

Replace each hexadecimal digit by its group of four bits.

s) $5B_{16}$

Hexadecimal digits	5	B
Groups of four bits	0101	1011

$5B_{16} = 0101101 1_{2}$

The leading zero may also be omitted: $101101 1_{2}$ .

Check via the decimal value: $5B_{16} = 5 \cdot 16 + 11 = 91$ and $101101 1_{2} = 64 + 16 + 8 + 2 + 1 = 91$ ✓

t) $8F2_{16}$

Hexadecimal digits	8	F	2
Groups of four bits	1000	1111	0010

$8F2_{16} = 10001111001 0_{2}$

Check via the decimal value: $8F2_{16} = 8 \cdot 256 + 15 \cdot 16 + 2 = 2290$ and $10001111001 0_{2} = 2048 + 128 + 64 + 32 + 16 + 2 = 2290$ ✓

Octal → hexadecimal

The way leads via binary: replace each octal digit by its three bits, then group the bits from the right in groups of four and replace each group by its hexadecimal digit.

u) $31 5_{8}$

Step 1: each octal digit becomes its group of three bits.

Octal digits	3	1	5
Groups of three bits	011	001	101

Step 2: without leading zeros, the bits are $1100110 1_{2}$ . Grouped from the right in groups of four:

Groups of four bits	1100	1101
Hexadecimal digits	C	D

$31 5_{8} = CD_{16}$

Check via the decimal value: $31 5_{8} = 3 \cdot 64 + 8 + 5 = 205$ and $CD_{16} = 12 \cdot 16 + 13 = 205$ ✓

v) $652 4_{8}$

Step 1: each octal digit becomes its group of three bits.

Octal digits	6	5	2	4
Groups of three bits	110	101	010	100

Step 2: without leading zeros, the bits are $11010101010 0_{2}$ . Grouped from the right in groups of four:

Groups of four bits	1101	0101	0100
Hexadecimal digits	D	5	4

$652 4_{8} = D54_{16}$

Check via the decimal value: $652 4_{8} = 6 \cdot 512 + 5 \cdot 64 + 2 \cdot 8 + 4 = 3412$ and $D54_{16} = 13 \cdot 256 + 5 \cdot 16 + 4 = 3412$ ✓

Hexadecimal → octal

The way leads via binary: replace each hexadecimal digit by its four bits, then group the bits from the right in groups of three and replace each group by its octal digit.

w) $E9_{16}$

Step 1: each hexadecimal digit becomes its group of four bits.

Hexadecimal digits	E	9
Groups of four bits	1110	1001

Step 2: without leading zeros, the bits are $1110100 1_{2}$ . Grouped from the right in groups of three, with one leading zero in the leftmost group:

Groups of three bits	011	101	001
Octal digits	3	5	1

$E9_{16} = 35 1_{8}$

Check via the decimal value: $E9_{16} = 14 \cdot 16 + 9 = 233$ and $35 1_{8} = 3 \cdot 64 + 5 \cdot 8 + 1 = 233$ ✓

x) $A7E_{16}$

Step 1: each hexadecimal digit becomes its group of four bits.

Hexadecimal digits	A	7	E
Groups of four bits	1010	0111	1110

Step 2: without leading zeros, the bits are $10100111111 0_{2}$ . Grouped from the right in groups of three:

Groups of three bits	101	001	111	110
Octal digits	5	1	7	6

$A7E_{16} = 517 6_{8}$

Check via the decimal value: $A7E_{16} = 10 \cdot 256 + 7 \cdot 16 + 14 = 2686$ and $517 6_{8} = 5 \cdot 512 + 64 + 7 \cdot 8 + 6 = 2686$ ✓

Wattsinside

Exercise DET-01.02 – Converting numbers between number systems (Solution)

Task

Solution

Given

Required

Binary → decimal

a) 111001012​

b) 1011010011012​

Octal → decimal

c) 2538​

d) 47268​

Hexadecimal → decimal

e) 9E16​

f) C3A16​

Decimal → binary

g) 18710​

h) 285410​

Decimal → octal

i) 20110​

j) 389010​

Decimal → hexadecimal

k) 22210​

l) 278810​

Binary → octal

m) 100111012​

n) 1110101100012​

Octal → binary

o) 3658​

p) 30428​

Binary → hexadecimal

q) 101001112​

r) 1101001110002​

Hexadecimal → binary

s) 5B16​

t) 8F216​

Octal → hexadecimal

u) 3158​

v) 65248​

Hexadecimal → octal

w) E916​

x) A7E16​

Table of Contents

Backlinks

a) $1110010 1_{2}$

b) $10110100110 1_{2}$

c) $25 3_{8}$

d) $472 6_{8}$

e) $9E_{16}$

f) $C3A_{16}$

g) $18 7_{10}$

h) $285 4_{10}$

i) $20 1_{10}$

j) $389 0_{10}$

k) $22 2_{10}$

l) $278 8_{10}$

m) $1001110 1_{2}$

n) $11101011000 1_{2}$

o) $36 5_{8}$

p) $304 2_{8}$

q) $1010011 1_{2}$

r) $11010011100 0_{2}$

s) $5B_{16}$

t) $8F2_{16}$

u) $31 5_{8}$

v) $652 4_{8}$

w) $E9_{16}$

x) $A7E_{16}$