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# Numeral Systems Calculator With Steps

## With this calculator you can convert integers and decimal numbers from one number system to another and get a detailed solution. Use a radix from 2 to 36.

 Number From 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 To 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 Number of digits after the decimal point (for decimal) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

Number Conversion Examples

Example 1

Let's convert the number 12 from the base-ten positional numeral system to the binary numeral system.

Convert the number 1210 to base-2, by successively dividing by 2, until the quotient is zero. The result will be the number formed from the remainders written from right to left.

 12 : 2 = 6 remainder: 0 6 : 2 = 3 remainder: 0 3 : 2 = 1 remainder: 1 1 : 2 = 0 remainder: 1

1210 = 11002

Example 2

Let's convert the number 12.3 from the base-ten positional numeral system to the binary numeral system.

Convert whole number part 12 of the number 12.310 to base-2, by successively dividing by 2, until the quotient is zero. The result will be the number formed from the remainders written from right to left.

 12 : 2 = 6 remainder: 0 6 : 2 = 3 remainder: 0 3 : 2 = 1 remainder: 1 1 : 2 = 0 remainder: 1

1210 = 11002

Let's convert the decimal part 0.3 of the number 12.310 to base-2, using successively multiplying by 2, until the decimal part of the product turns out to be zero or the required number of digits after the decimal point is reached. If after multiplying the whole number part is not zero, then you need to replace the value of the whole number part with zero. The result will be a number of integer parts of product written from left to right.

 0.3 · 2 = 0.6 0.6 · 2 = 1.2 0.2 · 2 = 0.4 0.4 · 2 = 0.8 0.8 · 2 = 1.6 0.6 · 2 = 1.2 0.2 · 2 = 0.4 0.4 · 2 = 0.8 0.8 · 2 = 1.6

0.310 = 0.0100110012
12.310 = 1100.0100110012

Example 3

Let's convert the number 10011 from the binary numeral system to the base-ten positional numeral system.

Let's convert the number 100112 to base-10 by first writing down the position of each digit in the number from right to left, starting at zero.

 Position 4 3 2 1 0 Number 1 0 0 1 1

Each digit position will be a power of 2, because the number system is base-2. It is necessary to write down the sum, each term in which will be the product of each digit of the number 100112 and 2 to the power, which is determined by the position in the number.

100112 = 1 ⋅ 24 + 0 ⋅ 23 + 0 ⋅ 22 + 1 ⋅ 21 + 1 ⋅ 20 = 1910

Example 4

Let's convert the number 11.101 from the binary numeral system to the base-ten positional numeral system.

Let's convert the number 11.1012 to base-10 by first writing down the position of each digit in the number

 Position 1 0 -1 -2 -3 Number 1 1 1 0 1

Each digit position will be a power of 2, because the number system is base-2. It is necessary to write down the sum, each term in which will be the product of each digit of the number 11.1012 and 2 to the power, which is determined by the position in the number.

11.1012 = 1 ⋅ 21 + 1 ⋅ 20 + 1 ⋅ 2-1 + 0 ⋅ 2-2 + 1 ⋅ 2-3 = 3.62510

Example 5

Let's convert the number 1583 from the base-ten positional numeral system to the hexadecimal numeral system.

Convert the number 158310 to base-16, by successively dividing by 16, until the quotient is zero. The result will be the number formed from the remainders written from right to left.

 1583 : 16 = 98 remainder: 15,   15 = F 98 : 16 = 6 remainder: 2 6 : 16 = 0 remainder: 6

158310 = 62F16

Example 6

Let's convert the number 1583.56 from the base-ten positional numeral system to the hexadecimal numeral system.

Convert whole number part 1583 of the number 1583.5610 to base-16, by successively dividing by 16, until the quotient is zero. The result will be the number formed from the remainders written from right to left.

 1583 : 16 = 98 remainder: 15,   15 = F 98 : 16 = 6 remainder: 2 6 : 16 = 0 remainder: 6

158310 = 62F16

Let's convert the decimal part 0.56 of the number 1583.5610 to base-16, using successively multiplying by 16, until the decimal part of the product turns out to be zero or the required number of digits after the decimal point is reached. If after multiplying the whole number part is not zero, then you need to replace the value of the whole number part with zero. The result will be a number of integer parts of product written from left to right.

 0.56 · 16 = 8.96 0.96 · 16 = 15.36,   15 = F 0.36 · 16 = 5.76 0.76 · 16 = 12.16,   12 = C 0.16 · 16 = 2.56 0.56 · 16 = 8.96 0.96 · 16 = 15.36,   15 = F 0.36 · 16 = 5.76 0.76 · 16 = 12.16,   12 = C 0.16 · 16 = 2.56

0.5610 = 0.8F5C28F5C216
1583.5610 = 62F.8F5C28F5C216= 62F.(8F5C2)16

Example 7

Let's convert the number A12DCF from the hexadecimal numeral system to the base-ten positional numeral system.
Let's convert the number A12DCF16 to base-10 by first writing down the position of each digit in the number from right to left, starting at zero.

 Position 5 4 3 2 1 0 Number A 1 2 D C F

Each digit position will be a power of 16, because the number system is base-16. It is necessary to write down the sum, each term in which will be the product of each digit of the number A12DCF16 and 16 to the power, which is determined by the position in the number.
A16 = 1010
D16 = 1310
C16 = 1210
F16 = 1510

A12DCF16 = 10 ⋅ 165 + 1 ⋅ 164 + 2 ⋅ 163 + 13 ⋅ 162 + 12 ⋅ 161 + 15 ⋅ 160 = 1056302310

Example 8

Let's convert the number A12DCF.12A from the hexadecimal numeral system to the base-ten positional numeral system.

Let's convert the number A12DCF.12A16 to base-10 by first writing down the position of each digit in the number

 Position 5 4 3 2 1 0 -1 -2 -3 Number A 1 2 D C F 1 2 A

Each digit position will be a power of 16, because the number system is base-16. It is necessary to write down the sum, each term in which will be the product of each digit of the number A12DCF.12A16 and 16 to the power, which is determined by the position in the number.
A16 = 1010
D16 = 1310
C16 = 1210
F16 = 1510

A12DCF.12A16 = 10 ⋅ 165 + 1 ⋅ 164 + 2 ⋅ 163 + 13 ⋅ 162 + 12 ⋅ 161 + 15 ⋅ 160 + 1 ⋅ 16-1 + 2 ⋅ 16-2 + 10 ⋅ 16-3 = 10563023.0727539062510

Example 9

Let's convert the number 1010100011 the binary numeral system to the the hexadecimal numeral system.

Let's convert the number 10101000112 to base-10 by first writing down the position of each digit in the number from right to left, starting at zero.

 Position 9 8 7 6 5 4 3 2 1 0 Number 1 0 1 0 1 0 0 0 1 1

Each digit position will be a power of 2, because the number system is base-2. It is necessary to write down the sum, each term in which will be the product of each digit of the number 10101000112 and 2 to the power, which is determined by the position in the number.

10101000112 = 1 ⋅ 29 + 0 ⋅ 28 + 1 ⋅ 27 + 0 ⋅ 26 + 1 ⋅ 25 + 0 ⋅ 24 + 0 ⋅ 23 + 0 ⋅ 22 + 1 ⋅ 21 + 1 ⋅ 20 = 67510

Convert the number 67510 to base-16, by successively dividing by 16, until the quotient is zero. The result will be the number formed from the remainders written from right to left.

 675 : 16 = 42 remainder: 3 42 : 16 = 2 remainder: 10,   10 = A 2 : 16 = 0 remainder: 2

67510 = 2A316
10101000112 = 2A316
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