## Quinary Number System

Subject: Compulsory Maths

#### Overview

Quinary Number System is a numeral system with 5 as the base.A possible origination of a quinary system is that there are five fingers on either hand. The base five is started from 0-4. In the quinary place system, five numerals, from 0 to 4, are used to represent any real number. According to this method, five is written as 10, twenty five is written as 100 and sixty is written as 220.

##### Quinary Number System

The quinary number system is a number system having five as the base. There are only five numerals in the quinary number system. They are 0, 1, 2, 3 and 4 in this system. This will represent any real numbers.

Quinary means base 5 so each place is a power of 5.

In this method five is written as 10, twenty-five is written as 100 and sixty is written as 220.

Consider the quinary number of 1555

1555 = 1 x 52 + 5 x51 + 5 x50

= 25 + 25 + 5

= 55

While converting a decimal number into a quinary number, we must divide it by 5 repeatedly and write the remainders until the result of the division is 0. The quinary number is obtained by reading the sequence of the remainders in the reverse order. For example, let's consider the number 8410

84 ÷ 5 = 16 Remainder 4

16 ÷ 5 = 3 Remainder 1

3÷ 5 = 0 Remainder 2

Finding arithmetic in a base other than 10 is to understand the notation we use in base 10.

We write the number thirteen as 13, meaning 1 tens and 3 ones. It may help you to think about objects, like sticks. The idea is to make thirteen sticks and arrange them in the group of ten. You get 1 groups of ten and three extra.

Suppose, if you add 23 and 19 you put together the 3 ones with the 9 ones giving 12 ones, which is 1 ten and 2 extra. That is you get one more group of ten sticks. That is the "carry over". So, altogether you have 2 + 1 + 1 tens and 2 ones, for a sum of 42.

In base-5, you want to collect the objects in groups of five rather than tens. So if you have nine objects you can arrange them into one group of five and 4 ones.

Now to add 2 and 3 using base 5 notation, 2 + 3 = 10 in base 5.

### Subtraction of quinary numbers

Subtraction in quinary number is straight forward as we are always subtracting a smaller digit from a large digit. Let's look at a base 10 problem first.

3 2 5

$\underline{-1 3 4}$

1 9 1

Starting in the right most column 5 - 4 = 1 but in the next column you need to borrow from the next column. Since this is base 10 notation you are borrowing ten so the 3 in the third column be 2 and adding to 10 to 2 you have 12 in the second column.

Now,

Let's try a base 5 problem

431

$\underline{-240}$

141

As in the base 10 problem, the first column is easy, 1 - 0 = 1. In the second you need to borrow from the third column. Since the numbers are written in base 5 notation you are borrowing five so the 4 in the third column becomes 3 and adding five to gives you eight in the second column.

Example:

Convert the following decimal number into a quinary number.

a) 425

Solution:

 5 425 0 5 85 0 5 17 2 5 3 3 0

∴ 42510 = 32005

##### Things to remember
• There are only five numerals 0, 1, 2, 3,  and 4 in quinary number system.
• Quinary means base 5 so each place is a power of 5.
• Many languages use the quinary number system.
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.

Solution:

 5 425 0 5 85 0 5 17 2 5 3 3

0

∴ 42510 = 32005

Solution:

 5 924 4 5 184 4 5 36 1 5 7 2 5 1 1 0

∴ 92410 = 121445

Solution:

 5 924 4 5 184 4 5 36 1 5 7 2 5 1 1 0

∴ 92410 = 121445

Solution:

 5 1574 4 5 312 2 5 62 2 5 12 2 5 2 2 0

∴ 157410 = 222245

Solution:

 5 2487 2 5 497 2 5 99 4 5 19 4 5 3 3 0

∴ 248710 = 344225

Solution:

 5 3040 0 5 608 3 5 121 1 5 24 4 5 4 4 0

∴ 304010 = 441305

Solution:

 5 5864 4 5 1172 2 5 234 4 5 46 1 5 9 5 5 1 1 0

∴ 586410 = 1514245

Solution:

1215

= 1 × 52 + 2× 51 + 1 × 50

= 25 + 10 + 1

= 3610

Then,

 2 36 0 2 18 0 2 9 1 2 4 0 2 2 0 2 1 1 0

∴ 3610 = 1001002

Solution:

2035

= 2× 52 + 0× 51 + 3× 50

= 50 + 0+ 3

= 5310

Then,

 2 53 1 2 26 0 2 13 1 2 6 0 2 3 1 1

∴ 5310 = 1101012

Solution:

4325

= 4× 52 + 3× 51 + 2× 50

= 100 + 15 + 2

= 15210

Then,

 2 152 0 2 76 0 2 38 0 2 19 1 2 9 1 2 4 0 2 2 0 1

∴ 15210 = 100110002

Solution:

1101102

= 1× 25 +1× 24+0× 23+1× 22+1× 21+0× 20

= 32 +16 +0 +4 +2 +0

= 5410

Now,

 5 54 4 5 10 0 5 2 2 0

∴ 1101102 = 5410 = 2045

Solution:

11101102

= 1× 26 + 1× 25 + 1 × 24 + 0× 23 + 1× 22 + 1× 21 + 0× 20

= 64 + 32 + 16 +0 + 4 + 2 +0

= 11810

Now,

 5 118 3 5 23 3 5 4 4 0

∴ 11101102 = 11810 = 4335

Solution:

111011012

= 1× 27 + 1× 26 + 1× 25 + 0× 24 + 1× 23 + 1×22 + 0× 21 + 1× 20

= 128 + 64 + 32 + 0+ 8 +4 +0 + 1

= 23710

Now,

 5 237 2 5 47 2 5 9 4 5 1 1 0

∴ 111011012 = 23710 = 14225

Solution:

Smallest possible number = 100002

Now,

100002 = (?)10

= 1× 24

= 1610

Then,

base5 =

 5 16 1 5 3 3 0

= 315

∴ 100002 = 1610 = 315

So, base10 = 1610

base 5 = 315

Solution:

43215

= 4× 53 + 3× 52 + 2× 51 + 1× 50

= 500 + 75 + 10 + 1

= 58610

Then,

 2 586 0 2 293 1 2 146 0 2 73 1 2 36 0 2 18 0 2 9 1 2 4 1 2 2 0 1

∴ 58610 = 10110010102

Solution:

33045

= 3× 53 + 3× 52 + 0× 51 + 4× 50

= 375 + 75 +0 + 4

= 45410

Then,

 2 454 0 2 227 1 2 113 1 2 56 0 2 28 0 2 14 0 2 7 1 2 3 1 1

∴ 45410 = 1110001102

Solution:

 5 1574 4 5 312 2 5 62 2 5 12 2 5 2 2 0

∴ 157410 = 222245

Solution:

 5 5864 4 5 1172 2 5 234 4 5 46 1 5 9 4 5 1 1 0

∴ 586410 = 1414245

Solution:

2125

= 2× 52 + 1× 51 + 2× 50

= 2× 25 + 1× 5 + 2× 1

= 50 + 5 + 2

= 5710

Solution:

23045

= 2× 53 + 3× 52 + 0× 41 + 4× 50

= 2× 125 + 3× 25 + 0×4 + 4×1

= 250 + 75 +0 +4

= 32910

Solution:

31045

= 3× 53 + 1× 52 + 0×51 + 4× 50

= 3× 125 + 1× 25 + 0×5 + 4×1

= 375 + 25 + 0+ 4

= 40410

Solution:

11102

= 1× 23 + 1× 22 + 1× 21+ 0×20

= 8 + 4 + 2 +0

= 1410

Now,

 5 14 4 5 2 2 0

∴ 11102 = 1410 = 245

Solution:

10112

= 1 × 23 + 0× 22 + 1× 21 + 1× 50

= 8 + 0 + 2 + 1

= 1110

Then,

 5 11 1 5 2 2 0

∴ 10112 = 1110 = 215

Solution:

11112

= 1× 23 + 1× 22 + 1× 21 + 1×20

= 8 + 4 + 2 + 1

= 1510

Then,

 5 15 0 5 3 3 0

∴ 11112 = 1510 = 305

Solution:

4415

= 4× 52+ 4× 51 + 1× 50

= 4× 25 + 4× 5 + 1

= 100 + 20 + 1

= 12110

Then,

 2 121 1 2 60 0 2 30 0 2 15 1 2 7 1 2 3 1 2 1 1 0

∴ 4415 = 12110 = 11110012

Solution:

21345

= 2× 53 + 1× 52 + 3× 51 + 4× 50

= 250 + 25 + 15 + 4

= 294

Then,

 2 244 0 2 122 0 2 61 1 2 30 0 2 15 1 2 7 1 2 3 1 2 1 1 0

∴ 29410 = 111101002