Calculator of the octal counting system subtraction. Calculator of surgery systems with solution. Jam units from the older discharge

With the help of this online calculator You can translate integers and fractional numbers from one number system to another. A detailed solution is given with explanations. To translate, enter the original number, set the source number system base, set the base of the number system to which you want to translate the number and click on the "Translate" button. Theoretical part and numerical examples see below.

The result is already received!

Translation of whole and fractional numbers from one number system to any other - theory, examples and solutions

There are positional and not positional number systems. Arabic number system that we use everyday life, It is a positional, and Roman - no. In positional surgery systems, the position of the number uniquely determines the value of the number. Consider this on the example of the number 6372 in a decimal number system. Number this number on the right left since scratch:

Then the number 6372 can be represented as follows:

6372 \u003d 6000 + 300 + 70 + 2 \u003d 6 · 10 3 + 3 · 10 2 + 7 · 10 1 + 2 · 10 0.

The number 10 defines the number system (in this case it is 10). As degrees, the positions of the number of this number are taken.

Consider real decimal number 1287.923. Number it starting from scratch the position of the number from the decimal point to the left and right:

Then the number 1287.923 can be represented as:

1287.923 \u003d 1000 + 200 + 80 + 7 + 0.9 + 0.02 + 0.003 \u003d 1 · 10 3 + 2 · 10 2 + 8 · 10 1 + 7 · 10 0 + 9 · 10 -1 + 2 · 10 -2 + 3 · 10 -3.

In general, the formula can be represented as follows:

C n · s. N + C N-1 · s. N-1 + ... + C 1 · s. 1 + C 0 · s 0 + d -1 · s -1 + d -2 · s -2 + ... + d -k · s -k

where c n is a number in position n., D -k - fractional number in position (-k), s. - Number system.

A few words about the number systems. The number in the decimal number system consists of a plurality of numbers (0.1,2,3,4,5,6,7,8,9), in an octaous number system - from a plurality of numbers (0.1, 2,3,4,5,6,7), in a binary number system - from a plurality of numbers (0.1), in a hexadecimal number system - from a plurality of numbers (0,1,2,3,4,5,6, 7,8,9, A, B, C, D, E, F), where A, B, C, D, E, F correspond to the number 10,11,12,13,14,15. In Table Table.1 The numbers are presented in different number systems.

Table 1
Notation
10 2 8 16
0 0 0 0
1 1 1 1
2 10 2 2
3 11 3 3
4 100 4 4
5 101 5 5
6 110 6 6
7 111 7 7
8 1000 10 8
9 1001 11 9
10 1010 12 A.
11 1011 13 B.
12 1100 14 C.
13 1101 15 D.
14 1110 16 E.
15 1111 17 F.

Translation of numbers from one number system to another

To transfer numbers from one number to another to another, the easiest way to first translate the number to a decimal number system, and then, from the decimal number system to translate to the desired number system.

Translation of numbers from any number system in a decimal number system

Using formula (1), you can translate numbers from any number system to a decimal number system.

Example 1. Translate the number 1011101.001 from the binary number system (SS) in a decimal SS. Decision:

1 · 2 6 +0 · 2 5 + 1 · 2 4 + 1 · 2 3 + 1 · 2 2 + 0 · 2 1 + 1 · 2 0 + 0 · 2 -1 + 0 · 2 -2 + 1 · 2 -3 \u003d 64 + 16 + 8 + 4 + 1 + 1/8 \u003d 93.125

Example2. Translate the number 1011101.001 from the octaous number system (SS) in a decimal SS. Decision:

Example 3 . Translate the number AB572.CDF from a hexadecimal number system in a decimal SS. Decision:

Here A. - per 10, B. - by 11, C.- at 12, F. - at 15.

Translation of numbers from a decimal number system to another number system

To transfer numbers from a decimal numbering system to another number system, it is necessary to translate separately by the integer part of the number and fractional part of the number.

An integer part of the number is translated from a decimal SS to another number system - a sequential division of a whole part of the number on the base of the number system (for a binary CC - by 2, for an 8-character SS - by 8, for 16-smoke-16, etc. ) Before getting a whole residue, less than the base of the SS.

Example 4 . We translate the number 159 of the decimal SS into the binary SS:

159 2
158 79 2
1 78 39 2
1 38 19 2
1 18 9 2
1 8 4 2
1 4 2 2
0 2 1
0

As can be seen from fig. 1, the number 159 during division by 2 gives the private 79 and the residue 1. Next, the number 79 during division by 2 gives Private 39 and the residue 1, etc. As a result, by building a number from the balances of divisions (right to left) we get a number in binary ss: 10011111 . Consequently, you can write:

159 10 =10011111 2 .

Example 5 . We translate the number 615 of the decimal SS into the octal SS.

615 8
608 76 8
7 72 9 8
4 8 1
1

When the number from the decimal SS in the octal SS, it is necessary to sequentially divide the number on 8 until the whole residue is less than 8. As a result, building a number from the balances of division (right to left), we get a number in the octane SS: 1147 (See Fig. 2). Consequently, you can write:

615 10 =1147 8 .

Example 6 . We transfer the number 19673 from the decimal number system to hexadecimal SS.

19673 16
19664 1229 16
9 1216 76 16
13 64 4
12

As can be seen from Fig.

To transfer the correct decimal fractions (real number with a zero integer) to the counting system with the base S, a given number must be multiplied to S, until a clean zero does not get in the fractional part, or we will not get the required number of discharges. If you get a number with a whole part, different from zero, then this whole part does not take into account (they are consistently enrolled in the result).

Consider the foregoing on the examples.

Example 7 . We transfer the number 0.214 from the decimal number system to binary SS.

0.214
x. 2
0 0.428
x. 2
0 0.856
x. 2
1 0.712
x. 2
1 0.424
x. 2
0 0.848
x. 2
1 0.696
x. 2
1 0.392

As can be seen from Fig. 4, the number 0.214 is multiplied by 2. If the multiplication is obtained with a whole part, different from zero, then the integer part is written separately (to the left of the number), and the number is written to the zero integer. If, when multiplying, a number with a zero integer is obtained, then zero is written to the left. The multiplication process continues until the fractional part does not get pure zero or do not get the required number of discharges. Recording fatty numbers (Fig. 4) from top to bottom We obtain the desired number in the binary number system: 0. 0011011 .

Consequently, you can write:

0.214 10 =0.0011011 2 .

Example 8 . We translate the number 0.125 from the decimal number system to binary SS.

0.125
x. 2
0 0.25
x. 2
0 0.5
x. 2
1 0.0

To bring the number of 0.125 of the decimal SS into a binary, this number is multiplied by 2. In the third stage it turned out 0. Therefore, the following result turned out:

0.125 10 =0.001 2 .

Example 9 . We translate the number 0.214 from the decimal number system to hexadecimal SS.

0.214
x. 16
3 0.424
x. 16
6 0.784
x. 16
12 0.544
x. 16
8 0.704
x. 16
11 0.264
x. 16
4 0.224

Following examples 4 and 5, we obtain numbers 3, 6, 12, 8, 11, 4. But in hexadecimal CC, the numbers 12 and 11 correspond to the number C and B. Therefore, we have:

0.214 10 \u003d 0.36C8B4 16.

Example 10 . We translate the number 0.512 from a decimal number system in the octal SS.

0.512
x. 8
4 0.096
x. 8
0 0.768
x. 8
6 0.144
x. 8
1 0.152
x. 8
1 0.216
x. 8
1 0.728

Received:

0.512 10 =0.406111 8 .

Example 11 . We translate the number 159.125 from a decimal number system to binary SS. To do this, we translate separately an integer part of the number (Example 4) and the fractional part of the number (Example 8). Next, we get the merging of these results:

159.125 10 =10011111.001 2 .

Example 12 . We transfer the number 19673.214 from a decimal number system to hexadecimal. To do this, we translate separately an integer part of the number (Example 6) and the fractional part of the number (example 9). Next, we get the combining results.

How do we add in the decimal number system?

Let's remember how we fold the numbers already familiar to us, in decimal.

The most important thing is to understand the discharge. Remember the alphabet of each SS and then it will be easier.

Addition in the binary system is no different from the addition in the decimal system. The main thing is to remember, the alphabet contains only two digits: 0 and 1. Therefore, when we fold 1 + 1, we get 0, and we increase the number for another 1 category. Look at the example above:

  1. We begin to fold as you used to right left. 0 + 0 \u003d 0, it means that you write 0. Go to the next discharge.
  2. We fold 1 + 1 and we obtain 2, but 2 is not in the binary number system, which means we write 0, and 1 add to the next discharge.
  3. We obtain in this discharge three units fold 1 + 1 + 1 \u003d 3, this figure can also be. So 3 - 2 \u003d 1. and 1 add to the next discharge.
  4. We again turn out 1 + 1 \u003d 2. We already know that 2 cannot be written 0, and 1 add to the next discharge.
  5. There is nothing more to fold, it means that we get: 10100.

We disassembled one example, the second decide on your own:

As well as in any other number systems, it is necessary to remember the alphabet. Let's try to fold the expression.

  1. All as usual, we begin to fold on the right left. 4 + 3 \u003d 7.
  2. 5 + 4 \u003d 9. Nine can not be, it means from 9 subtract 8, we get 1. And 1 add to the next discharge.
  3. 3 + 7 + 1 \u003d 11. From 11, we subtract 8, we get 3. and add to the next discharge.
  4. 6 + 1 = 7.
  5. There is nothing to fold. Answer: 7317.

Now do the addition independently:

  1. We carry out already familiar to us actions and do not forget about the alphabet. 2 + 1 \u003d 3.
  2. 5 + 9 \u003d 14. Remember the alphabet: 14 \u003d E.
  3. C \u003d 12. 12 + 8 \u003d 20. Twenty no in a hexadecimal number system. So, we deduct 16 from 20 and we get 4. and add to the next discharge.
  4. 1 + 1 = 2.
  5. There is nothing more to fold. Answer: 24E3.

Deduction in number systems

Recall how we do it in a decimal number system.

  1. We start from left to right, from a smaller discharge to more. 2 - 1 \u003d 1.
  2. 1 – 0 = 1.
  3. 3 - 9 \u003d? Troika less than nine, so we conscate the unit from the older discharge. 13 - 9 \u003d 4.
  4. From the last discharge, we took a unit for the previous action, therefore 4 - 1 \u003d 3.
  5. Answer: 3411.

  1. We start as usual. 1 - 1 \u003d 0.
  2. 1 – 0 = 1.
  3. From 0 to take away a unit. Therefore, we take one discharge from the elder. 2 - 1 \u003d 1.
  4. Answer: 110.

Now decide independently:

  1. Nothing new, the main thing is to remember the alphabet. 4 - 3 \u003d 1.
  2. 5 – 0 = 5.
  3. From 3 to take away 7, we cannot immediately, for this, we need to borrow a unit from an older discharge. 11 - 7 \u003d 4.
  4. We remember that they borrowed a unit earlier, 6 - 1 \u003d 5.
  5. Answer: 5451.

Take the previous example, and let's see what will be the result in the hexadecimal system. Same or other?

  1. 4 – 3 = 1.
  2. 5 – 0 = 5.
  3. From 3 to take away 7, we cannot immediately, for this, we need to borrow a unit from an older discharge. 19 - 7 \u003d 12. In the hexadecimal system 12 \u003d C.
  4. We remember that they borrowed a unit earlier, 6 - 1 \u003d 5
  5. Answer: 5C51

An example for an independent solution:

Multiplication in number systems

Let's remember once and for all that multiplication in any number system per unit will always give the same number.

  1. Every discharge is multiplied by one, as usually right left, and we obtain the number 6748;
  2. 6748 multiply on 8 and get the number 53984;
  3. We proceed the operation of multiplication 6748 by 3. We obtain the number 20244;
  4. We fold all 3 numbers by the rules. We get 2570988;
  5. Answer: 2570988.

In the binary system multiplied very easily. We always multiply either on 0 or by one. The main thing is to carefully fold. Let's try.

  1. 1101 multiply by one, as usually right left, and we obtain the number 1101;
  2. We produce this operation 2 more times;
  3. We fold all the 3 numbers carefully, remember about the alphabet, not forgetting about the ladder;
  4. Answer: 1011011.

An example for an independent solution:

  1. 5 x 4 \u003d 20. A 20 \u003d 2 x 8 + 4. The balance of division is written to the number - it will be 4, and 2 hold in the mind. We do this procedure right to left and obtain the number 40234;
  2. When multiplying by 0, we get four 0;
  3. When multiplying by 7, we turn out the number 55164;
  4. Now we add numbers and get - 5556634;
  5. Answer: 5556634.

An example for an independent solution:

Everything as usual, the main thing to remember the alphabet. Letter digits, for convenience, translate into the usual number system, as multiply, translate back to the letter value.

Let's look at the 5th 20A4 multiplication by visibility.

  1. 5 x 4 \u003d 20. A 20 \u003d 16 + 4. The balance of division is written to the number - it will be 4, and 1 is kept in the mind.
  2. And x 5 + 1 \u003d 10 x 5 + 1 \u003d 51. 51 \u003d 16 x 3 + 3. The remainder of the division is written to the number - it will be 3, and 3 kept in the mind.
  3. With multiplication by 0, we obtain 0 + 3 \u003d 3;
  4. 2 x 5 \u003d 10 \u003d a; As a result, we obtain A334; We do this procedure with two other numbers;
  5. Remember the multiplication rule by 1;
  6. When multiplying on B, we obtain the number 1670s;
  7. Now we add numbers and get - 169V974;
  8. Answer: 169В974.

An example for an independent solution.

Consider the main arithmetic operations: addition, subtraction, multiplication and division. The rules for performing these operations in the decimal system are well known - this is an addition, subtraction, multiplication of a column and an angle division. These rules apply to all other positional surgery. It is only necessary to use special folding tables and multiplication for each system.

1. Addition

The folding tables are easy to compile using the account rules.

When adding, the figures are summed up by discharge, and if excess occurs, it is transferred to the left.

Example 1. Moving the number 15 and 6 in various number systems.

Example 2. Moving the number 15, 7 and 3.

Hexadecimal : F 16 +7 16 +3 16

15+7+3 = 25 10 = 11001 2 = 31 8 = 19 16 .

Check:

11001 2 = 2 4 + 2 3 + 2 0 = 16+8+1=25,

31 8 = 3 . 8 1 + 1 . 8 0 = 24 + 1 = 25,

19 16 = 1 . 16 1 + 9 . 16 0 = 16+9 = 25.

Example 3. Moving numbers 141.5 and 59.75.

Answer: 141.5 + 59.75 \u003d 201.25 10 \u003d 11001001,01 2 \u003d 311.2 8 \u003d C9,4 16

Check. We transform the amount received to decimal:

11001001,01 2 = 2 7 + 2 6 + 2 3 + 2 0 + 2 -2 = 201,25

311,2 8 = 3 . 8 2 + 1 . 8 1 + 1 . 8 0 + 2 . 8 -1 = 201,25

C9,4 16 \u003d 12 . 16 1 + 9 . 16 0 + 4 . 16 -1 = 201,25

2. Subtraction

Subtraction in a binary number system

minuend

subtrahend

0

1

0

1

loan

Subtraction in a hexadecimal number system

0

1

2

3

4

5

6

7

8

9

A.

B.

C.

D.

E.

F.

0

1

2

3

4

5

6

7

8

9

A.

B.

C.

D.

E.

F.

Jam units from the older discharge

Subtraction in the octal number system

0

1

2

3

4

5

6

7

0

1

2

3

4

5

6

7

Loanunits of senior discharge

Example 4. Subscribe a unit from numbers 10 2 , 10 8 and 10. 16

Example 5. Submount the unit from numbers 100 2 , 100 8 and 100. 16 .

Example 6. Pull out the number 59.75 from among 201.25.

Answer: 201.25 10 - 59.75 10 \u003d 141.5 10 \u003d 10001101.1 2 \u003d 215.4 8 \u003d 8D, 8 16.

Check. We transform the obtained differences to the decimal form:

10001101,1 2 = 2 7 + 2 3 + 2 2 + 2 0 + 2 -1 = 141,5;

215,4 8 = 2 . 8 2 + 1 . 8 1 + 5 . 8 0 + 4 . 8 -1 = 141,5;

8D, 8 16 \u003d 8 . 16 1 + D . 16 0 + 8 . 16 -1 = 141,5.

Examples of translation numbers in various systems Note

Example №1
Translate the number 12 of the decimal in the binary number system
Decision

We transfer the number 12 10 to a 2-imaging system of number, with the help of a sequential division by 2, until the incomplete private will not be zero. As a result, a number from the separation remains recorded on the right left will be obtained.

12 : 2 = 6 Residue: 0.
6 : 2 = 3 Residue: 0.
3 : 2 = 1 Residue: 1.
1 : 2 = 0 Residue: 1.

12 10 = 1100 2

Example number 2.
We transfer the number 12.3 of the decimal to the binary number system

12.3 10 = 1100.010011001100110011001100110011 2

Decision

We transfer the whole part 12 of the number 12.3 10 into a 2-imaging system of number, with the help of a sequential division by 2, until the incomplete private will be zero. As a result, a number from the separation remains recorded on the right left will be obtained.

12 : 2 = 6 Residue: 0.
6 : 2 = 3 Residue: 0.
3 : 2 = 1 Residue: 1.
1 : 2 = 0 Residue: 1.

12 10 = 1100 2

We translate the fractional part 0.3 of the number 12.3 10 to a 2-imaging system of the number, using a sequential multiplication by 2, until it turns out to be zero in the fractional part of the work, or the required number of semicolons will not be reached. If, as a result of multiplication, the whole part is not zero, then it is necessary to replace the value of the integer on zero. As a result, a number from integer parts of the works recorded from left to right will be obtained 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
0.6 · 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
0.6 · 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
0.6 · 2 = 1 .2
0.2 · 2 = 0 .4
0.4 · 2 = 0 .8
0.8 · 2 = 1 .6
0.6 · 2 = 1 .2

0.3 10 = 0.010011001100110011001100110011 2
12.3 10 = 1100.010011001100110011001100110011 2

Example number 3.
We translate the number 10011 from the binary system in a decimal number system
Decision

We transfer the number 10011 2 to a decimal number system, to do this, we first write down the position of each figure including with the right to left, starting from zero

Each position of the digit will be the degree of number 2, since the number is 2-чеч. It is necessary to consistently multiply each number 10011 2 to 2 to the degree of the corresponding position of the number and then folded with the subsequent product of the following number to the degree of corresponding position.

10011 2 = 1 ⋅ 2 4 + 0 ⋅ 2 3 + 0 ⋅ 2 2 + 1 ⋅ 2 1 + 1 ⋅ 2 0 = 19 10

Example number 4.
We translate the number 11.101 from the binary system to a decimal number system

11.101 2 = 3.625 10

Decision

We translate the number 11.101 2 to a decimal number system, for this, first write down the position of each figure in the number

Each position of the digit will be the degree of number 2, since the number is 2-чеч. It is necessary to consistently multiply each number 11.101 2 to 2 to the degree of corresponding position of the number and then folded with the subsequent product of the next number to the degree of its corresponding position.

11.101 2 = 1 ⋅ 2 1 + 1 ⋅ 2 0 + 1 ⋅ 2 -1 + 0 ⋅ 2 -2 + 1 ⋅ 2 -3 = 3.625 10

Example number 5.
We translate the number 1583 of the decimal system in a hexadecimal number system

1583 10 \u003d 62F 16

Decision

We translate the number 1583 10 to the 16-imaging system of the number, with the help of a sequential division by 16, until the incomplete private will not be zero. As a result, a number from the separation remains recorded on the right left will be obtained.

1583 : 16 = 98 Residue: 15, 15 \u003d F
98 : 16 = 6 Residue: 2.
6 : 16 = 0 Rest: 6.

1583 10 \u003d 62F 16

Example number 6.
We translate the number 1583.56 from the decimal system in a hexadecimal system

1583.56 10 \u003d 62F.8F5C28F5C28F5C28F5C28F5C28F5C2 16

Decision

We transfer the whole part of 1583 of the number 1583.56 10 into a 16-imaging system of the number, with the help of a sequential division by 16, until the incomplete private will be zero. As a result, a number from the separation remains recorded on the right left will be obtained.

1583 : 16 = 98 Residue: 15, 15 \u003d F
98 : 16 = 6 Residue: 2.
6 : 16 = 0 Rest: 6.

1583 10 \u003d 62F 16

We translate a fractional part of 0.56 of the number 1583.56 10 into a 16-imaging system of the number, with the help of a sequential multiplication by 16, until in the fractional part of the work, it is not possible to get zero or the required number of semicolons will not be reached. If, as a result of multiplication, the whole part is not zero, then it is necessary to replace the value of the integer on zero. As a result, a number from integer parts of the works recorded from left to right will be obtained from left to right.

0.56 · 16 = 8 .96
0.96 · 16 = 15.36, 15 \u003d F
0.36 · 16 = 5 .76
0.76 · 16 = 12.16, 12 \u003d C
0.16 · 16 = 2 .56
0.56 · 16 = 8 .96
0.96 · 16 = 15.36, 15 \u003d F
0.36 · 16 = 5 .76
0.76 · 16 = 12.16, 12 \u003d C
0.16 · 16 = 2 .56
0.56 · 16 = 8 .96
0.96 · 16 = 15.36, 15 \u003d F
0.36 · 16 = 5 .76
0.76 · 16 = 12.16, 12 \u003d C
0.16 · 16 = 2 .56
0.56 · 16 = 8 .96
0.96 · 16 = 15.36, 15 \u003d F
0.36 · 16 = 5 .76
0.76 · 16 = 12.16, 12 \u003d C
0.16 · 16 = 2 .56
0.56 · 16 = 8 .96
0.96 · 16 = 15.36, 15 \u003d F
0.36 · 16 = 5 .76
0.76 · 16 = 12.16, 12 \u003d C
0.16 · 16 = 2 .56
0.56 · 16 = 8 .96
0.96 · 16 = 15.36, 15 \u003d F
0.36 · 16 = 5 .76
0.76 · 16 = 12.16, 12 \u003d C
0.16 · 16 = 2 .56

0.56 10 \u003d 0.8F5C28F5C28F5C28F5C28F5C28F5C2 16
1583.56 10 \u003d 62F.8F5C28F5C28F5C28F5C28F5C28F5C2 16

Example number 7.
We translate the number A12DCF from a hexadecimal system to a decimal number system

A12DCF 16 \u003d 10563023 10

Decision

We translate the number A12DCF 16 to a decimal number system, for this, first write down the position of each figure including from the right to left, starting from scratch

Each number of numbers will be the degree of number 16, since the number is 16-ich. It is necessary to multiply each number A12DCF 16 to 16 to the degree of the corresponding position of the number and then folded with the subsequent product of the following number to the degree of its corresponding position.
2

1 0 -1 -2 -3 NumberA.1 2 D.C.F.1 2 A.
Each number of numbers will be the degree of number 16, since the number is 16-ich. It is necessary to consistently multiply each number A12DCF.12A 16 to 16 to the degree of the corresponding position of the number and then folded with the subsequent product of the following number to the degree of its corresponding position.
A 16 \u003d 10 10
D 16 \u003d 13 10
C 16 \u003d 12 10
F 16 \u003d 15 10

A12DCF.12A 16 \u003d 10 ⋅ 16 5 + 1 ⋅ 16 4 + 2 ⋅ 16 3 + 13 ⋅ 16 2 + 12 ⋅ 16 1 + 15 ⋅ 16 0 + 1 ⋅ 16 -1

1 0 Number1 0 1 0 1 0 0 0 1 1
Each position of the digit will be the degree of number 2, since the number is 2-чеч. It is necessary to consistently multiply each number 1010100011 2 to 2 to the degree of corresponding position of the number and then folded with the subsequent product of the following number to the degree of corresponding position.

1010100011 2 = 1 ⋅ 2 9 + 0 ⋅ 2 8 + 1 ⋅ 2 7 + 0 ⋅ 2 6 + 1 ⋅ 2 5 + 0 ⋅ 2 4 + 0 ⋅ 2 3 + 0 ⋅ 2 2 + 1 ⋅ 2 1 + 1 ⋅ 2 0 = 675 10

We translate the number 675 10 to a 16-imaging number system, using a sequential division by 16, until the incomplete private will be zero. As a result, a number from the separation remains recorded on the right left will be obtained.

675 : 16 = 42 Residue: 3.
42 : 16 = 2 residue: 10, 10 \u003d a
2 : 16 = 0 Residue: 2.

675 10 \u003d 2A3 16