Addition in the binary number system is carried out according to the rules

0 + 0 = 0 | 0 + 1 = 1 | 1 + 0 = 1 | 1 + 1 u003d 2 _{10} u003d 10 _{2} (one goes to the most significant digit) |

The subtraction table in binary number system has the form

0 – 0 = 0 | 1 – 0 = 1 | 1 – 1 = 0 | 0 – 1 = 10 _{2} |
10 _{2} – 1 u003d 1 (we take the unit from the highest digit) |

The multiplication table in the binary system has the form

0 x 0 = 0 | 0 x 1 = 0 | 1 x 0 = 0 | 1 x 1 = 1 |

The division table in the binary system has the form

0 : 0 undefined | 1 : 0 undefined | 0 : 1 = 0 | 1 : 1 = 1 |

The reverse code of a number in a system with base *p* is a number in this system, obtained by replacing a digit, a symbol in each bit of a number with its complement to the maximum digit in the system (that is, up to *p* – 1).

Additional code u003d reverse code + one in the least significant digit.

**Example.**

1. 10011 Þ binary number,

01100 Þ the return code of this binary number,

01101 Þ additional code of this binary number;

2. 457 Þ octal number,

321 Þ additional code;

3. A9 Þ hexadecimal number,

57 Þ additional code.

Subtraction using two’s complement: find the two’s complement of the subtrahend of the same bit length as the minuend, and add this code to the minuend. The result of the subtraction will be the amount received without taking into account the most significant digit (discarded).

**Example.** Let’s perform the subtraction directly and through addition (via two’s complement):

**EXAMPLES OF TYPICAL SOLUTIONS**

**Task #1**

*Given two numbers a=D7 _{16} and b=331 _{8} . It is necessary to determine which of the numbers written in the binary system satisfies the inequality a*

1) 11011001 _{2}

2) 11011100 _{2}

3) 11010111 _{2}

4) 11011000 _{2}

General approach: translate all numbers (both initial data and answers) into one (any!) number system and compare.

__Solution (option 1, via decimal):__

one)

2)

3) We translate all the answers into the decimal system:

11011001 _{2} = 217,

11011100 _{2} = 220,

11010111 _{2} = 215,

11011000 _{2} =216

4) Obviously, between the numbers 215 and 217 there can only be 216

5) Thus, the correct answer is 4.

__Solution (option 2, via binary):__

one) (each hexadecimal digit is *separately* translated into four binary – *tetrad* );

2) (each digit of the octal system is *separately* translated into three binary – *triad* , leading zeros can be omitted);

3) Now you need to figure out that between these numbers there is only a binary number 11011000 _{2} – this is the answer 4.

__Solution (option 3, via octal):__

one) (first transferred to the binary system, then the binary notation of the number was divided into triads **from right to left** , each triad was transferred *separately* to the decimal system, since for numbers from 0 to 7 their octal notation coincides with the decimal notation);

2) , there is no need to translate anywhere;

3) We translate all the answers into the octal system:

11011001 _{2} u003d 011 011 001 _{2} u003d 331 _{8} (divided into triads **from right to left** , each triad was transferred *separately* to the decimal system, as in paragraph 1)

11011100 _{2} = 334 _{8} , 11010111 _{2} = 327 _{8} , 11011000 _{2} = 330 _{8}

4) In the octal system, between the numbers 327 _{8} and 331 _{8} there can only be 330 _{8}

5) Thus, the correct answer is 4.

__Solution (option 4, via hexadecimal system):__

one) there is no need to translate;

2) (first transferred to the binary system, then the binary notation of the number was divided into tetrads **from right to left** , each tetrad was transferred to the hexadecimal system; while tetrads can be converted from binary to *decimal,* and then replace all numbers greater than 9 with letters – A, B , C, D, E, F);

3) We translate all the answers into hexadecimal system:

11011001 _{2} u003d 1101 1001 _{2} u003d D9 _{16} (broken into tetrads **from right to left** , each tetrad was transferred *separately* to the decimal system, all numbers greater than 9 were replaced with letters – A, B, C, D, E, F, as in p. one)

11011100 _{2} = DC _{16} , 11010111 _{2} = D7 _{16} , 11011000 _{2} = D8 _{16}

4) In the hexadecimal system, between the numbers D7 _{16} and D9 _{16} there can only be D8 _{16}

5) Thus, the correct answer is 4.

**Task #2**

Given two numbers x and y. What is the sum of these numbers if and ?

1) 121 _{8}

2) 171 _{8}

3) 69 _{16}

4) 1000001 _{2}

General approach: translate both original numbers and answers into one (any!) number system, and perform addition

__Solution (option 1, via decimal):__

one.

2.

3. Addition: 35 + 86 = 121.

4 *a* ) we translate the result into all systems in which answers are given (until we find the right one):

121 = 1111001 _{2} = 171 _{8} = 79 _{16}

4 *b* ) or we translate all answers into a decimal system

121 _{8} = 81,

171 _{8} = 121,

69 _{16} = 105,

1000001 _{2} = 65.

5) So the correct answer is 2.

__Solution (option 2, via binary):__

one) (each digit of the octal system is *separately* translated into three binary – *triad* , leading zeros can be omitted).

2) (each hexadecimal digit is *separately* translated into four binary – *tetrad* ).

3) Add up

**100011 _{2}**

**+ 1010110 _{2 .}**

**1111001 _{2}**

4) Convert all answers to the binary system

121 _{8} u003d 001 010 001 _{2} u003d 1010001 _{2} (by triads)

171 _{8} u003d 001 111 001 _{2} u003d 1111001 _{2} (by triads)

69 _{16} u003d 0110 1001 _{2} u003d 1101001 _{2} (by tetrads)

1000001 _{2} does not need to be translated.

5) The correct answer is 2.

__Solution (option 3, via octal):__

one) , no need to translate

2) (at first

transferred to the binary system, then the binary notation of the number was divided into triads **from right to left** , each triad was transferred *separately* to the decimal system, since for numbers from 0 to 7 their octal notation coincides with the decimal)

3) Add up

**43 _{8}**

**+ 126 _{8} .**

**171 _{8}**

4) We see that there is such an answer, this is answer 2.

__Solution (option 4, via hexadecimal system):__

one) (first transferred to the binary system, then the binary notation of the number was divided into tetrads **from right to left** , each tetrad was transferred to the hexadecimal system; while tetrads can be converted from binary to *decimal,* and then replace all numbers greater than 9 with letters – A, B , C, D, E, F).

2) , there is no need to translate.

3) Add up

**23 _{16}**

**+ 56 _{16} .**

**79 _{16}**

4) We translate all the answers into hexadecimal system:

121 _{8} u003d 001 010 001 _{2} u003d 0101 0001 _{2} u003d 51 _{16} (transferred to the binary system by triads, divided into tetrads **from right to left** , each tetrad was transferred *separately* to the decimal system, all numbers greater than 9 were replaced with letters – A, B, C, D, E, F).

171 _{2} = 001 111 001 _{2} = 0111 1001 _{2} = 79 _{16} ,

69 _{16} , no need to translate

1000001 _{2} = 0100 0001 _{2} = 41 _{16} .

5) So the correct answer is 2.

**Task #3**

**Example 1.** *Indicate, separated by commas, in ascending order, all decimal numbers not exceeding* 25 *, the record of which in the base four number system ends in* 11 *?*

General Approach:

– recall the algorithm for converting a number from a decimal system to a system with a base , it follows that the least significant digit of the result is the remainder of dividing the original number by , and the two lower digits are the remainder of the division by etc.;

– in this case , the remainder of dividing the number by must be equal to 11 _{4} = 5;

– therefore, the task is reduced to determining all numbers that are less than or equal to 25 and give a remainder of 5 when divided by 16.

__Solution (option 1, via decimal):__

1. General form of numbers that give a remainder of 5 when divided by 16:

where – non-negative integer (0, 1, 2, …)

2. Among all such numbers, you need to choose those that are less than or equal to 25 (“do not exceed 25”); there are only two of them: 5 (with ) and 21 (at ).

3. Thus, the correct answer is 5, 21.

__Solution (option 2, through the quaternary system, proposed by O.A. Tuzova):__

1) Let’s translate 25 into the quaternary number system: 25 = 121 _{4} , all numbers of interest to us are not greater than this value.

2) Of these numbers, we select only those that end in 11, there are only two such numbers:

it’s 11 _{4} = 5 and 111 _{4} = 21

3) Thus, the correct answer is 5, 21.

**Example 2.** *Indicate, separated by commas, in ascending order, all the bases of the number systems in which the entry of the number* 23 *ends in* 2 *.*

General Approach:

– here is the inverse problem – the base of the number system is unknown, we will denote it by *N* ;

– since the last digit of the number is 2, the base must be greater than 2, that is, *N* > 2;

– recall the algorithm for converting a number from a decimal system to a system with a base (see presentation), it follows that the least significant digit of the result is the remainder of dividing the original number by *N.*

Decision:

1) So, you need to find all integers , such that the remainder after dividing 23 by is 2, or (which is the same)

(*)

where is a non-negative integer (0, 1, 2, …).

2) The difficulty is that , and unknown, but here you need to “play” on the fact that these are *natural numbers* .

3) From the formula (*) we obtain , so the problem is to find all divisors of 21 that are greater than 2.

4) There are only three such divisors in this problem: and 21.

5) Thus, the correct answer is 3, 7, 21.

**Task number 4.**

*Indicate, separated by commas, in ascending order, all the bases of the number systems in which the entry for the number* 31 *ends in* 11 *.*

General Approach:

– the base of the number system is unknown, we will denote it by

– for now we will assume that the entry of the number 31 in the system with the base consists of three digits, and the two lower ones (11) are given to us, and one (we denote it by ) Need to find:

2 1 0 ← digits

31 u003d k 1 1 _{N} u003d k N ^{2} + N ^{1} + N ^{0} u003d k N ^{2} + N + 1;

– it can be shown that with a larger number of digits this formula is also true, that is, the number 31 can be represented as for some integer ; for example, for a number with five digits, we get:

4 3 2 1 0 ← digits

31 = k _{4} k _{3} k _{2} 1 1 _{N} = k _{4} N ^{4} + k _{3} N ^{3} + k _{2} N ^{2} + N ^{1} + N ^{0}

= k N ^{2} + N + 1

for (from the first three terms the common factor was taken out ).

Decision:

1) So, you need to find all integers , such that

(**)

where is a non-negative integer (0, 1, 2, …).

2) The difficulty is that , and unknown, but here you need to “play” on the fact that these are *natural numbers* .

3) From the formula (**) we obtain , so the problem is to find all the divisors number 30 and select only those for which the equation (**) is solvable for an integer , i.e, is an integer.

4) Write out all the divisors of the number 30 greater than or equal to 2: 2, 3, 5, 6, 10, 15, 30.

5) Of all these divisors, only 2, 3, 5 and 30 have a value is an integer (it is equal to 7, 3, 1 and 0, respectively).

6) Thus, the correct answer is 2, 3, 5, 30.

**INDIVIDUAL TASKS**

**Task #1**

**Option number 1**

How is the number _{8310} represented in the binary number system?

1) 1001011 _{2}

2) 1100101 _{2}

3) 1010011 _{2}

4) 101001 _{2}

**Option number 2**

How many ones are there in binary notation for the number 195?

fifteen

2) 2

3) 3

4) 4

**Option number 3**

How many ones are there in binary notation for the number 173?

1) 7

2) 5

3) 6

4) 4

**Option number 4**

How is the number 25 represented in binary?

1) 1001 _{2}

2) 11001 _{2}

3) 10011 _{2}

4) 11010 _{2}

**Option number 5**

How is the number 82 represented in binary?

1) 1010010 _{2}

2) 1010011 _{2}

3) 100101 _{2}

4) 1000100 _{2}

**Option number 6**

How is the number 263 represented in the octal number system?

1) 301 _{8}

2) 650 _{8}

3) 407 _{8}

4) 777 _{8}

**Option number 7**

How is the number 567 _{8} written in the binary system?

1) 1011101 _{2}

2) 100110111 _{2}

3) 101110111 _{2}

4) 11110111 _{2}

**Option number 8**

How is the number A87 _{16} written in the octal number system?

1) 435 _{8}

2) 1577 _{8}

3) 5207 _{8}

4) 6400 _{8}

**Option number 9**

How is the number 754 _{8} written in hexadecimal notation?

1) 738 _{16}

2) 1A4 _{16}

3) 1EC _{16}

4) A56 _{16}

**Option number 10**

One byte is used to store a signed integer. How many units does the internal representation of the number **(-128)** contain?

eleven

2) 2

3) 3

4) 4

**Option number 11**

One byte is used to store a signed integer. How many units does the internal representation of the number **(-35)** contain?

thirteen

2) 4

3) 5

4) 6

**Option number 12**

Given: , . Which of the numbers C, written in the binary system, satisfies the inequality ?

1) 10011010

2) 10011110

3) 10011111

4) 11011110

**Option number 13**

Given: , . Which of the numbers C, written in the binary system, satisfies the inequality ?

1) 11111001

2) 11011000

3) 11110111

4) 11111000

**Option number 14**

Given: , . Which of the numbers C, written in the binary system, satisfies the inequality ?

1) 11011010

2) 11111110

3) 11011110

4) 11011111

**Option number 15**

Given: , . Which of the numbers C, written in the binary system, satisfies the inequality ?

1) 11101010

2) 11101110

3) 11101011

4) 11101100

**Option number 16**

Given: , . Which of the numbers C, written in the binary system, satisfies the inequality ?

1) 11101010

2) 11101000

3) 11101011

4) 11101100

**Option number 17**

Given: *a* u003d D7 _{16} , *b* u003d 331 _{8} . Which of the numbers *c* , written in the binary system, meets the condition *a* < *c* < *b* ?

1) 11011001

2) 11011100

3) 11010111

4) 11011000

**Option number 18**

Given: , . Which of the numbers C, written in the binary system, satisfies the inequality ?

1) 11010011

2) 11001110

3) 11001010

4) 11001100

**Option number 19**

Given: , . Which of the numbers C, written in the binary system, satisfies the inequality ?

1) 10011010

2) 10011110

3) 10011111

4) 11011110

**Option number 20**

How is the number _{18910} represented in binary number system?

1) 1001011 _{2}

2) 1100101 _{2}

3) 1010011 _{2}

4) 10111101 _{2}

**Task #2**

**Option number 1**

Calculate the sum of numbers *x* and *y* , for *x =* A6 _{16} , *y =* 75 _{8} . Present the result in binary number system.

1) 11011011 _{2}

2) 11110001 _{2}

3) 11100011 _{2}

4) 10010011 _{2}

**Option number 2**

The value of the expression 10 _{16} + 10 _{8} • 10 _{2} in binary is

1) 1010 _{2}

2) 11010 _{2}

3) 100000 _{2}

4) 110000 _{2}

**Option number 3**

Calculate the sum of binary numbers *x* and *y* if *x* = 1010101 _{2} and *y* = 1010011 _{2}

1) 10100010 _{2}

2) 10101000 _{2}

3) 10100100 _{2}

4) 10111000 _{2}

**Option number 4**

Calculate the value of the sum 10 _{2} + 10 _{8} +10 _{16} in binary.

1) 10100010 _{2}

2) 11110 _{2}

3) 11010 _{2}

4) 10100 _{2}

**Option number 5**

Calculate the sum of numbers *x* and *y* , for *x =* 271 _{8} , *y =* 11110100 _{2} . Express the result in hexadecimal number system.

1) 151 _{16}

2) 1AD _{16}

3) 412 _{16}

4) 10B _{16}

**Option number 6**

Calculate the sum of the numbers *x* and *y* , with *x =* A1 _{16} , *y =* 1101 _{2} . Express the result in decimal notation.

1) 204

2) 152

3) 183

4) 174

**Option number 7**

Calculate the sum of numbers *x* and *y* , for *x =* 56 _{8} , *y =* 1101001 _{2} . Present the result in binary number system.

1) 11110111 _{2}

2) 10010111 _{2}

3) 1000111 _{2}

4) 11001100 _{2}

**Option number 8**

Calculate the sum of numbers *x* and *y* , for *x =* 5A _{16} , *y =* 1010111 _{2} . Express the result in octal number system.

1) 151 _{8}

2) 261 _{8}

3) 433 _{8}

4) 702 _{8}

**Option number 9**

Calculate the sum of numbers *x* and *y* , for *x =* 127 _{8} , *y =* 10010111 _{2} . Express the result in decimal notation.

1) 214

2) 238

3) 183

4) 313

**Option number 10**

Calculate A81 _{16} + 377 _{16} . Present the result in the same number system.

1) 21B _{16}

2) DF8 _{16}

3) C92 _{16}

4) F46 _{16}

**Option number 11**

What is the difference between the numbers 101 _{16} and 110111 _{2} ?

1) 312 _{8}

2) 12 _{8}

3) 32 _{16}

4) 64 _{16}

**Option number 12**

What is the difference between the numbers 124 _{8} and 52 _{16} ?

1) 11 _{2}

2) 10 _{2}

3) 100 _{2}

4) 110 _{2}

**Option number 13**

What is the sum of the numbers _{278} and _{3416} ?

1) 113 _{8}

2) 63 _{8}

3) 51 _{16}

4) 110011 _{2}

**Option number 14**

What is the sum of the numbers 43 _{8} and 56 _{16} ?

1) 79 _{16}

2) A3 _{16}

3) 125 _{8}

4) 1010101 _{2}

**Option number 15**

What is the sum of the numbers 43 _{8} and 56 _{16} ?

1) 121 _{8}

2) 171 _{8}

3) 69 _{16}

4) 1000001 _{2}

**Option number 16**

Calculate the sum of numbers X and Y if X=110111 _{2} Y=135 _{8} ^{.} Express the result in binary form.

1) 11010100 _{2}

2) 10100100 _{2}

3) 10010011 _{2}

4) 10010100 _{2}

**Option number 17**

Calculate the sum of numbers *x* and *y* , with *x =* B8 _{16} , *y =* 77 _{8} . Present the result in binary number system.

1) 11011011 _{2}

2) 11110001 _{2}

3) 11110111 _{2}

4) 10010011 _{2}

**Option number 18**

Calculate the sum of numbers *x* and *y* , with *x =* F6 _{16} , *y =* 35 _{8} . Present the result in binary number system.

1) 11011011 _{2}

2) 11110001 _{2}

3) 11100011 _{2}

4) 100010011 _{2}

**Option number 19**

What is the sum of the numbers _{538} and _{6616} ?

1) 127 _{8}

2) 372 _{8}

3) 69 _{16}

4) 10010001 _{2}

**Option number 20**

What is the difference between the numbers 1101 _{16} and 1101111 _{2} ?

1) 312 _{8}

2) 122 _{8}

3) 32A _{16}

4) 1170 _{16}

**Task #3**

**Option number 1**

Indicate, separated by commas, in ascending order, all the bases of the number systems in which the entry of the number 22 ends in 4.

**Option number 2**

In a number system with some base, the number 12 is written as 110. Specify this base.

**Option number 3**

Indicate, separated by commas, in ascending order, all the bases of the number systems in which the entry of the number 39 ends in 3.

**Option number 4**

Indicate, separated by commas, in ascending order, all the bases of the number systems in which the entry of the number 29 ends in 5.

**Option number 5**

In a number system with some base, the decimal number 129 is written as 1004. Specify this base.

**Option number 6**

Indicate, separated by commas, in ascending order, all the bases of the number systems in which the entry of the number 40 ends in 4.

**Option number 7**

In a number system with some base, the decimal number 25 is written as 100. Find this base.

**Option number 8**

Indicate, separated by commas, in ascending order, all the bases of the number systems in which the entry of the number 27 ends in 3.

**Option number 9**

Indicate, separated by commas, in ascending order, all decimal numbers not exceeding 26, the record of which in the ternary number system ends in 22?

**Option number 10**

Indicate, separated by commas, in ascending order, all decimal numbers not exceeding 30, the record of which in the quaternary number system ends in 31?

**Option number 11**

Indicate, separated by a comma, in ascending order, all decimal natural numbers not exceeding 17, the record of which in the ternary number system ends in two identical digits?

**Option number 12**

Indicate how many times the number 3 occurs in the notation of the numbers 19, 20, 21, …, 33 in the number system with base 6.

**Option number 13**

Indicate how many times the number 1 occurs in the notation of the numbers 12, 13, 14, …, 31 in the number system with base 5.

**Option number 14**

Indicate, separated by commas, in ascending order, all the bases of the number systems in which the entry of the number 23 ends in 1.

**Option number 15**

Indicate, separated by commas, in ascending order, all the bases of the number systems in which the entry of the number 63 ends in 23.

**Option number 16**

Specify, separated by commas, in ascending order, all decimal numbers not exceeding 25, the notation of which in the base four number system ends in 11.

**Option number 17**

In a number system with some base, the decimal number 49 is written as 100. Specify this base.

**Option number 18**

Indicate, separated by commas, in ascending order, all the bases of the number systems in which the entry of the number 29 ends in 5.

**Option number 19**

In a number system with some base, the decimal number 129 is written as 1004. Find this base.

**Option number 20**

Indicate how many times the number 2 occurs in the notation of the numbers 10,11,12, …, 17 in the number system with base 5.

**Task #4**

Convert the given number from decimal to binary, octal and hexadecimal.

**Option number 1**

a) 860 _{(10)} ;

b) 785 _{(10)} ;

c) 149.375 _{(10)} ;

d) 953.25 _{(10)} ;

e) 228.79 _{(10)} .

**Option number 2**

a) 250 _{(10)} ;

b) 757 _{(10)} ;

c) 711.25 _{(10)} ;

d) 914.625 _{(10)} ;

e) 261.78 _{(10)} .

**Option number 3**

a) 759 _{(10)} ;

b) 265 _{(10)} ;

c) 79.4375 _{(10)} ;

d) 360.25 _{(10)} ;

e) 240.25 _{(10)} .

**Option number 4**

a) 216 _{(10)} ;

b) 336 _{(10)} ;

c) 741.125 _{(10)} ;

d) 712.375 _{(10)} ;

e) 184.14 _{((10)} .

**Option number 5**

a) 530 _{(10)} ;

b) 265 _{(10)} ;

c) 597.25 _{(10)} ;

d) 300.375 _{(10)} ;

e) 75.57 _{(10)} .

**Option number 6**

a) 945 _{(10)} ;

b) 85 _{(10)} ;

c) 444.125 _{(10)} ;

d) 989.375 _{(10)} ;

e) 237.73 _{(10)} .

**Option number 7**

a) 287 _{(10)} ;

b) 220 _{(10)} ;

c) 332.1875 _{(10)} ;

d) 652.625 _{(10)} ;

e) 315.21 _{(10)} .

**Option number 8**

a) 485 _{(10)} ;

b) 970 _{(10)} ;

c) 426.375 _{(10)} ;

d) 725.625 _{(10)} ;

e) 169.93 _{(10)} .

**Option number 9**

a) 639 _{(10)} ;

b) 485 _{(10)} ;

c) 581.25 _{(10)} ;

d) 673.5 _{(10)} ;

e) 296.33 _{(10)} .

**Option number 10**

a) 618 _{(10)} ;

b) 556 _{(10)} ;

c) 129.25 _{(10)} ;

d) 928.25 _{(} _{10)} ;

e) 155.45 _{(10)} .

**Option number 11**

a) 772 _{(10)} ;

b) 71 _{(10)} ;

c) 284.375 _{(10)} ;

d) 876.5 _{(10)} ;

e) 281.86 _{(10)} .

**Option number 12**

a) 233 _{(10)} ;

b) 243 _{(10)} ;

c) 830.375 _{(10)} ;

d) 212.5 _{(10)} ;

e) 58.89 _{(10).}

**Option number 13**

a) 218 _{(10)} ;

b) 767 _{(10)} ;

c) 894.5 _{(10)} ;

d) 667.125 _{(10)} ;

e) 3.67 _{(10)} .

**Option number 14**

a) 898 _{(10)} ;

b) 751 _{(10)} ;

c) 327.375 _{(10)} ;

d) 256.625 _{(10)} ;

e) 184.4 _{(10)} .

**Option number 15**

a) 557 _{(10)} ;

b) 730 _{(10)} ;

c) 494.25 _{(10)} ;

d) 737.625 _{(10)} ;

e) 165.37 _{(10)} .

**Option number 16**

a) 737 _{(10)} ;

6) 92 _{(10)} ;

c) 934.25 _{(10)} ;

d) 413.5625 _{(10)} ;

e) 100.94 _{(10)} .

**Option number 17**

a) 575 _{(10)} ;

b) 748 _{(10)} ;

c) 933.5 _{(10)} ;

d) 1005.375 _{(10)} ;

e) 270.44 _{(10)} .

**Option number 18**

a) 563 _{(10)} ;

b) 130 _{(10)} ;

c) 892.5 _{(10)} ;

d) 619.25 _{(10)} ;

e) 198.05 _{(10)} .

**Option number 19**

a) 453 _{(10)} ;

b) 481 _{(10)} ;

c) 461.25 _{(10)} ;

d) 667.25 _{(10)} ;

e) 305.88 _{(10)} .

**Option number 20**

a) 949 _{(10)} ;

b) 763 _{(10)} ;

c) 994.125 _{(10)} ;

d) 523.25 _{(10)} ;

e) 203.82 _{(10)} .

**Task number 5**

Convert this number to decimal.

**Option number 1**

a) 1001010 _{(2)} ;

b) 1100111 _{(2)} ;

c) 110101101.00011 _{(2)} ;

d) 111111100.0001 _{(2)} ;

e) 775.11 _{(8)} ;

e) 294.3 _{(16).}

**Option number 2**

a) 1111000 _{(2)} ;

b) 1111000000 _{(2)} ;

c) 111101100.01101 _{(2)} ;

d) 100111100.1101 _{(2)} ;

e) 1233.5 _{(8)} ;

f) 2B3,F4 _{(} _{I6)} .

**Option number 3**

a) 1001101 _{(2)} ;

b) 10001000 _{(2)} ;

c) 100111001.01 _{(2)} ;

d) 1111010000.001 _{(2)} ;

e) 1461.15( _{8} );

f) 9D,A _{(16)}

**Option number 4**

a) 1100000110 _{(2)} ;

b) 1100010 _{(2)} ;

c) 1011010.001 _{(2)} ;

d) 10101000.001 _{(2)} ;

e) 1537.22( _{8} );

f) 2D9.8 _{(16)} .

**Option number 5**

a) 101000111 _{(2)} ;

b) 110001001 _{(2)} ;

c) 1001101010.01 _{(2)} ;

d) 1011110100.01 _{(2)} ;

e) 1317.75( _{8} );

f) 2F4.0С _{(16)} .

**Option number 6**

a) 110001111 _{(2)} ;

b) 111010001 _{(2)} ;

c) 100110101.001 _{(2)} ;

d) 10000010.01011 _{(2)} ;

e) 176.5( _{8} );

f) 3D2.04 _{(16)} .

**Option number 7**

a) 10101000 _{(2)} ;

b) 1101100 _{(2)} ;

c) 10000010000.01001 _{(2)} ;

d) 1110010100.001 _{(2)} ;

e) 1714.2( _{8} ); f) DD,3 _{(16)} .

**Option number 8**

a) 10101000 _{(2)} ;

b) 101111110 _{(2)} ;

c) 1010101.101 _{(2)} ;

d) 1111001110.01 _{(2)} ;

e) 721.2( _{8} );

f) 3C9.8 _{(16)} .

**Option number 9**

a) 1011000011 _{(2)} ;

b) 100010111 _{(2)} ;

c) 1100101101.1 _{(2)} ;

d) 1000000000.01 _{(2)} ;

e) 1046.4( _{8} );

e) 388.64 _{(16)} .

**Option number 10**

a) 1000001111 _{(2)} ;

b) 1010000110 _{(2)} ;

c) 101100110.011011 _{(2)} ;

d) 100100110.101011 _{(2)} ;

e) 10232.2 _{(8)} ;

e) 53.9 _{(16)} .

**Option number 11**

a) 1001101111 _{(2)} ;

b) 1000001110 _{(2)} ;

c) 111110011.011 _{(2)} ;

d) 11010101.1001 _{(2)} ;

e) 1634.5 _{(8)} ;

f) C2.3 _{(16)} .

**Option number 12**

a) 1111100010 _{(2)} ;

b) 1000011110 _{(2)} ;

c) 101100001.011101 _{(2)} ;

d) 1001111001.1 _{(2)} ;

e) 1071.54 _{(8)} ;

f) 18B,0C _{(16)} .

**Option number 13**

a) 101110100 _{(2)} ;

b) 1111101101 _{(2)} ;

c) 1110100001.01 _{(2)} ;

d) 1011111010.0001 _{(2)} ;

e) 744.12 _{(8)} ;

f) IE,C _{(16)} .

**Option number 14**

a) 101001101 _{(2)} ;

b) 1110111100 _{(2)} ;

c) 10000001000.001 _{(2)} ;

d) 1000110110.11011 _{(2)} ;

e) 147.56 _{(8)} ;

f) 1CA,3 _{(16)} .

**Option number 15**

a) 1110000010 _{(2)} ;

6) 1000100 _{(2)} ;

c) 110000100.001 _{(2)} ;

d) 1001011111.00011 _{(2)} ;

e) 665.42 _{(8)} ;

f) 246.18(, _{6} ).

**Option number 16**

a) 1010000 _{(2)} ;

b) 10010000 _{(2)} ;

c) 1111010000.01 _{(2)} ;

d) 101000011.01 _{(2)} ;

e) 1004.1 _{(8)} ;

e) 103.8C _{(16)} .

**Option number 17**

a) 11100001 _{(2)} ;

b) 101110111 _{(2)} ;

c) 1011110010.0001 _{(2)} ;

d) 1100010101.010101 _{(2} );

e) 533.2 _{(8)} ;

f) 32.22 _{(16)} .

**Option number 18**

a) 111001010 _{(2)} ;

b) 1101110001 _{(2)} ;

c) 1001010100.10001 _{(2)} ;

d) 111111110.11001 _{(2)} ;

e) 1634.35 _{(8)} ;

e) 6B,A _{(16)} .

**Option number 19**

a) 1110001111 _{(2)} ;

b) 100011011 _{(2)} ;

c) 1001100101.1001 _{(2)} ;

d) 1001001.011 _{(2)} ;

e) 335.7 _{(8)} ;

f) 14C,A _{(16)} .

**Option number 20**

a) 1100010010 _{(2)} ;

b) 10011011 _{(2)} ;

c) 1111000001.01 _{(2)} ;

d) 10110111.01 _{(2)} ;

e) 416.1(c>; f) 215.7 _{(16)} .

**Task number 6**

Perform arithmetic operations.

**Option number 1**

one.

a) 1101100000 _{(2)} + 10110110 _{(2)} ;

b) 101110111 _{(2)} + 1000100001 _{(2)} ;

c) 1001000111.01 _{(2)} + 100001101.101 _{(2)} ;

d) 271.34 _{(8)} + 1566.2 _{(8)} ;

e) 65.2 _{(16)} + ZSA.8 _{(16)} .

2.

a) 1011001001 _{(2)} -1000111011 _{(2)} ;

b) 1110000110 _{(2)} -101111101 _{(2)} ;

c) 101010000.10111 _{(2)} -11001100.01 _{(2)} ;

d) 731.6( _{8} ) – 622.6 _{(8)} ;

e) 22D,l _{(} _{l6)} -123.8 _{(16)} .

3.

a) 1011001 _{(2)} x 1011011 _{(2)} ;

b) 723.l _{(8)} x 50.2 _{(8)} ;

c) 69.4 _{(16)} x A, B _{(16).}

**Option number 2**

one.

a) 1010101 _{(2)} + 10000101 _{(2)} ;

b) 1111011101 _{(2)} + 101101000 _{(2)} ;

c) 100100111.001 _{(2)} + 100111010.101 _{(2)} ;

d) 607.54 _{(8)} + 1620.2 _{(8)} ;

e) 3BF,A _{(16)} + 313,A _{(16)} .

2.

a) 1001000011 _{(2)} -10110111 _{(2)} ;

b) 111011100 _{(2)} -10010100 _{(2)} ;

c) 1100110110.0011 _{(2)} -11111110.01(2);

d) 1360.14 _{(8)} -1216.4 _{(8)} ;

e) 33B.6 _{(16)} – 11B.4 _{(16)} .

3.

a) 11001 _{(2)} x 1011100 _{(2)} ;

b) 451.2 _{(8)} x 5.24 _{(8)} ;

c) 2B, A _{(16)} x 36.6 _{(16)} .

*Option number 3*

one.

a) 100101011 _{(2)} + 111010011 _{(2);}

b) 1001101110 _{(2)} + 1101100111 _{(2)} ;

c) 1010000100.1 _{(2)} + 11011110.001 _{(2)} ;

d) 674.34 _{(8)} + 1205.2 _{(8)} ;

e) 2FE,6 _{(16)} + ZV,4 _{(16).}

2.

a) 1100110010 _{(2)} – 1001101101 _{(2)} ;

b) 1110001100 _{(2)} -10001111 _{(2)} ;

c) 11001010.01 _{(2)} -1110001.001 _{(2)} ;

d) 641.6 _{(8)} – 273.04 _{(8)}

e) ZSE, B8 _{(16)} – 39A, B8 _{(16).}

3.

a) 1010101 _{(2)} _{*} 1011001( _{2)} ;

b) 1702.2 _{(8) *} 64.2 _{(8)} ;

c) 7.4 _{(16)*} 1D4 _{(16)}

**Option number 4**

one.

a) 101111111 _{(2)} + 1101110011 _{(2);}

b) 10111110 _{(2)} + 100011100 _{(2)} ;

c) 1101100011.0111 _{(2)} + 1100011.01 _{(2)} ;

d) 666.2 _{(8)} + 1234.24 _{(8)} ;

e) 346.4 _{(16)} + ZF2.6 _{(16)} .

2.

a) 1010101101 _{(2)} – 110011110 _{(2)} ;

b) 1010001111 _{(2)} -1001001110 _{(2)} ;

c) 1111100100.0111 _{(2)} -101110111.011 _{(2)} ;

d) 1437.24 _{(8)} – 473.4 _{(8);}

e) 24A.4 ( _{16)} – B3.8 _{(16)} .

3.

a) 101011 _{(2)} _{*} 100111( _{2)} ;

b) 1732.4 _{(8) *} 34.5 _{(8)} ;

c) 36.4 _{(16) *} A, A _{(16)} .

**Option number 5**

one.

a) 1100011010 _{(2)} + 11101100 _{(2);}

b) 10111010 _{(2)} + 1010110100 _{(2)} ;

c) 1000110111.011 _{(2)} + 1110001111.001 _{(2)} ;

d) 1745.5 _{(8)} + 1473.2 _{(8)} ;

e) 24D.5 _{(16)} + 141.4 _{(16)} .

2.

a) 1100101010 _{(2)} – 110110010 _{(2)} ;

b) 110110100 _{(2)} -110010100 _{(2)} ;

c) 1101111111.1 _{(2)} -1100111110.011 _{(2)} ;

d) 1431.26 _{(8)} – 1040.3 _{(8);}

e) 22C.6 _{(16)} – 54.2 _{(16)} .

3.

a) 1001001 _{(2)} _{*} 11001 _{(2)} ;

b) 245.04 _{(8) *} 112.2 _{(8)} ;

c) 4B,2 _{(16)*} 3C,3 _{(16)} .

**Option number 6**

one.

a) 10000011101 _{(2)} + 1010000010 _{(2);}

b) 100000001 _{(2)} + 1000101001 _{(2)} ;

c) 101111011.01 _{(2)} + 1000100.101 _{(2)} ;

d) 1532.14 _{(8)} + 730.16 _{(8)} ;

e) BB,4 _{(16)} + 2F0,6 _{(16)} .

2.

a) 1000101110 _{(2)} – 1111111 _{(2)} ;

b) 1011101000 _{(2)} -1001000000 _{(2)} ;

c) 1000101001.1 _{(2)} -1111101.1 _{(2)} ;

d) 1265.2 _{(8)} – 610.2 _{(8);}

e) 409.D (16 _{)} – 270.4 _{(16)} .

3.

a) 111010 _{(2)} _{*} 1100000 _{(2)} ;

b) 1005.5 _{(8) *} 63.3 _{(8)} ;

c) 4А,3 _{(16)*} F,6 _{(16)} .

**Option number 7**

one.

a) 1100110 _{(2)} + 1011000110 _{(2);}

b) 1000110 _{(2)} + 1001101111 _{(2)} ;

c) 101001100.101 _{(2)} + 1001001100.01 _{(2)} ;

d) 275.2 _{(8)} + 724.2 _{(8)} ;

e) 165.6 _{(16)} + 3E,B _{(16)} .

2.

a) 1011111111 _{(2)} – 100000011 _{(2)} ;

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