The sum of the digit terms 2. The sum of the digit terms of a natural number. What is multiplication

§1. The concept of "bit terms"

In this lesson, we will get acquainted with the concept of "bit terms" and learn how to decompose numbers into bit terms.

Let's solve the problem:

Little Red Riding Hood went to visit her grandmother.

And she took with her a present for her grandmother - a basket of pies.

Little Red Riding Hood had 10 cabbage pies and 7 mushroom pies in her basket. How many pies does Little Red Riding Hood have in the basket?

To answer the question of the problem, it is necessary to perform addition, namely, to 10 pies with cabbage add 7 pies with mushrooms.

10 + 7 = 17 (pies).

This means that there were 17 pies in total in Little Red Riding Hood's basket.

Let's pay attention to the numerical expression obtained when solving the problem:

Let's name all the components of the addition.

The first number 10 is the first term, the number 7 is the second term and the number 17 is the sum.

What else can we say about the numbers 10, 7 and 17?

The number 10 is a two-digit number written in two digits 1 and 0.

The number 10 belongs to the tens place and equals 1 ten.

The number 7 is a one-digit number written in one digit 7.

This number belongs to the category of ones.

Let's replace the terms 10 and 7 in our numerical expression with digit numbers.

So, the first term is 10 = 1 ten, and the second term is 7 = 7 units.

We got the following numeric expression:

1 dozen + 7 units = 17.

This means that the number 17 is a two-digit number written in two digits 1 and 7.

It consists of 1 dozen and 7 units.

Let's pay attention to the resulting expression: 1 dozen + 7 units = 17.

Let's call the components of addition.

The first term is 1 dozen, the second term is 7 units, the sum is the number 17.

Both the first and second terms are represented by bit numbers.

Hence, these terms can be called bit terms.

§2. Decomposition of numbers into bit terms

Let's write the numeric expressions 10 + 7 = 17 and 1 tens + 7 units = 17 as one numeric expression:

1 dozen + 7 units = 10 + 7 = 17.

The terms 10 and 7 will also be bit terms, so 10 = 1 ten, and 7 = 7 ones.

For example, the number 53 consists of 5 tens and 3 ones.

53 = 5 tens + 3 units = 50 + 3

Representation of a number in the form: 53 = 50 + 3 is called decomposition of a number into bit terms or the sum of bit terms.

And the numbers 50 and 3 are called bit terms.

Numbers 1, 10, 100, 1000, etc. - are called bit units.

So, 1 is a unit of the ones category;

10 - tens unit;

100 is a unit of the category of hundreds, etc.

For example, about the number 50, we can say that it is 5 units of the tens place, and about the number 3, we can say that these are 3 units of the ones place.

1.determine the number of all units of any category, i.e. how many units, tens, hundreds, etc .;

2. write down the number as a sum of bit terms.

Let's represent another number, the number 72, in the form of bit terms:

Let us underline units in this number with one line, and tens with two lines: 72.

Let's write the number 72 as the sum of the bit terms.

§3. Lesson summary

Let's summarize the lesson:

Any positive integer number can be represented as a sum of bit terms.

Representation of a number in the form: 53 = 50 + 3 is called the decomposition of the number into bit terms or the sum of bit terms. And the numbers 50 and 3 are called bit terms.

To decompose a number into bit terms, you must:

1) determine the number of all units of any category, i.e. how many units, tens, hundreds, etc .;

2) write down the number as a sum of bit terms.

Numbers 1, 10, 100, 1000, etc. - are called bit units. So, 1 is a unit of the ones category; 10 - tens unit; 100 - a unit of the category of hundreds, etc.

SOURCES

https://vimeo.com/124205288

http://znaika.ru/catalog/2-klass/matematika/Razryadnye-slagaemye

Summary of a lesson in mathematics.

Class: 2 class "B".

Teacher: Bukhteeva I.M.

Theme: A three-digit number as the sum of bit terms.

Lesson Objectives:

Further study of the bit (positional) principle of numbering three-digit numbers;

The procedure for decomposing a number into bit terms (the sum of the bit terms of a three-digit number);

Recognition of the bit composition of a number by its short decimal notation;

Formation of UUD: self-test according to the sample, communicative UUD (pair work).

Propaedeutics: addition and subtraction of three-digit numbers.

Repetition: "Round" numbers, bit terms.

Methods and techniques for organizing student activities:explanation of new material on tasks and illustrations of the textbook with a phased inclusion of students in independent activities; verbal counting.

Educational and didactic support:U-2, T-2, Z., models of the number 100, colored and simple pencils, pointer.

During the classes:

1. Organizing time.

Greetings from the teacher. Preparation of workplaces. Inclusion in the business rhythm of the lesson.

1. Updating students' knowledge.

• We repeat the sixth column of TU along the chain.

1. Lesson topic message. Goal setting.

• We suggest opening the textbook on p. 15, read the topic of the lesson ("Three-digit number as the sum of digit terms") and name any three-digit number.
• What will we learn in the lesson?

1. Statement of the educational problem.

Task number 1 (U-2, p. 15)

* We ask the students to look at the drawing of three patterns of the number 100 and answer the questions: how many cells are colored red? (200) In blue? (50) Yellow? (eight)

Explain while writing on the board.

Filled:

200 + 50 + 8 cells, which is equal to the number 258.

200 + 50 + 8 is the sum of the digit terms of the number 258, since this is 2 hundred. +5 dec. + 8 units (hundreds place, tens place and ones place).

After all the numbers are written as a sum of bit terms, we check the solutions by writing on the board under the dictation of the children:

258 - 200 + 50 + 8 1 65 = 100 + 60 + 5

319 = 300 +10 + 9 689 = 600 + 80 + 9 940 = 900 + 40 + 0

208 = 200 + 0 + 8 208 = 200 + 0 + 8 = 200 + 8

• We draw the children's attention to the bit terms - 940 = 900 + 40 + 0 and 208 = 200 + 0 + 8 - and explain that these sums of bit terms can be written differently: 940 - 900 + 40; 208 = 200 + 8, omitting the digit 0 in the bit terms.
• We carry out the second part of the task. We name the bit terms of each of the numbers,starting from the rank of hundreds, for example:

bit terms of the number 258. The rank of hundreds - 2 hundred., the rank of tens - 5 woods, the rank of units - 8;

digit terms of the number 208. The rank of hundreds - 2 hundred., the rank of tens - 0 dec, the rank of units - 8.

1. Primary anchoring.

Task number 3 (U-2, p. 16)

• Students read the assignment on their own and orally name the numbers that Masha missed (141, 146).
• We pay special attention to the wording “no more than 9 units”, explaining that the number 149 includes 1 hundred, 4 tens and 9 units. The number of units here is 9, that is, no more than 9.
• We ask the children to write down in a notebook all the numbers in order, in which 3 hundred., 5 dess. and no more than 7 units.
• We give time to complete the task, after which we conduct an oral check (350, 351, 352 ... 357).

Task number 4 (U-2, p. 16)

• Children perform the task verbally.
• Pupils, as a rule, do not name the number 340. It is advisable to clarify that the uncertainty in the category of ones ("several units") also allows the number 340 to be indicated, where the number of ones is written in the number 0: 340 is 3 hundreds and another 4 tens, and a few more units that are equal to 0.

Task number 5 (U-2, p. 16) has a combinatorial nature and refers to tasks of increased difficulty

• We invite students to read the assignment on their own and make three-digit numbers from such bit terms as 500 and 800, 40 and 70, 3 and 9.
• We give time for an independent search, and then we propose a solution algorithm based on fixing the bit term of the most significant bit and manipulating the bit terms of the least significant bit:
• 543, 549, 843, 849 (students add the missing numbers - 573, 579, 873, 879).

Task number 6 (U-2, p. 16)

We give students time to complete the task on their own and ask: why equality 437= 400 + 37 cannot be called the sum of the digit terms? (The tens place and the ones place are not highlighted.)

We propose to convert this equality into the sum of the digit terms and write on the board:

437 = 400 + 30 + 7

1. Independent work with verification against the standard.

Task number 1 (T-2, p. 7)

• Students read and complete the assignment independently,
• We ask the children, following the pattern written on the board, to check, by exchanging notebooks, the correctness of the assignment:

643 = 600 + 40 + 3 999 = 900 + 90 + 9 207 = 200+ 7
910 = 900 4 10 207 = 200 + 7 909 = 900 + 9

We identify the presence of errors, analyze each of them.

As a rule, errors occur in cases where the bit terms are written as 0: 910 = 900 + 10:

207 = 200 + 7: 909 = 900 + 9 .

We explain that the records: 910 = 900 + 10 and 910 = 900 + 10 + 0, 207 = 207 = 200 + 0 + 7, 909 = 900 + 9 and 909 = 900 + 0 + 9 are equal.

The bit term, which is denoted by the number 0, mathematics is not written. But if you write down the digit with the number 0, showing that in the tens place - 0 tens or in the ones - 0 ones, then there will be no error.

Task number 2 (T-2, p. 7)

Students read and complete the assignment independently.

Task number 3 (T-2, p. 7) Task 1

• Students read the problem on their own. We ask you to underline the key words of the condition in red pencil (“taken out 500 cents”, “there are 200 cents less left”), and in blue - the keywords of the requirement (“How many centners”, “left”).
• We read aloud the keywords of the condition and answer the requirement of the problem - we are looking fora value that is less than 500 centners per 200 centners:

500 q - 200 q = 300 q Answer: 300 q left.

• We ask: is it possible to find out how many centners of vegetables were in the warehouse?
• We write a short statement of the new problem on the board, askdecide on your ownand write down the answer.

They took out 500 c

300 c 500 c + 300 c = 800 c left Answer: 800 c was.

Assignment for home: repeat the seventh column of the Multiplication Table; No. 3, task 2and No. 4 (T-2, p. 7); cut a rectangle (13 cm * 8 cm) from a sheet of clean paper.Tasks that were not completed in the lesson.

1. Reflection of activity.

They are all different. For example, 2, 67, 354, 1009. Let's consider these numbers in detail.
2 consists of one digit, therefore such a number is called, single digit... Another example of single-digit numbers: 3, 5, 8.
67 consists of two digits, so this number is called, two-digit number... An example of two-digit numbers: 12, 35, 99.
Three-digit numbers consist of three digits, for example: 354, 444, 780.
Four-digit numbers consist of four digits, for example: 1009, 2600, 5732.

Two-digit, three-digit, four-digit, five-digit, six-digit, etc. numbers are called, multi-digit numbers.

Digits of numbers.

Consider the number 134. Each digit of this number has its place. Such places are called discharges.

Digit 4 takes place or place of ones. The number 4 can also be called a number. the first category.
The number 3 takes the place or place of the tens. Or the number 3 can be called a number second category.
And the number 1 is in the hundreds place. In another way, the number 1 can be called a number third category. Digit 1 is the last digit of the glory of the number 134, so digit 1 can be called the highest digit. The highest digit is always greater than 0.

Every 10 units of any rank form a new unit of a higher rank. 10 units form one tens place, 10 tens form one hundreds place, ten hundred form a thousand place, etc.
If there is no bit, then instead of it there will be 0.

For example: number 208.
Digit 8 is the first digit of ones.
Digit 0 is the second digit of tens. 0 means nothing in mathematics. It follows from the record that this number does not have tens.
Number 2 is the third place of hundreds.

This parsing of a number is called bit composition of the number.

Classes.

Multi-digit numbers are divided into groups of three digits from right to left. Such groups of numbers are called classes. The first class on the right is called class of units, the second is called class of thousands, third - class of millions, fourth - class of billions, fifth - trillion class, the sixth - class quadrillion, seventh - class quintillion, eighth - class sextillion.

Unit class- the first class to the right from the end of the three digits consists of the ones place, the tens place and the hundreds place.
Thousand class- the second class consists of a category: units of thousands, tens of thousands and hundreds of thousands.
Class of millions- the third class consists of a category: units of millions, tens of millions and hundreds of millions.

Let's look at an example:
We have the number 13 562 006 891.
This number has 891 units in the unit class, 6 units in the thousand class, 562 units in the million class, and 13 units in the billion class.

13 billion 562 million 6 thousand 891.

The sum of the bit terms.

Anyone with different digits can be decomposed into the sum of the bit terms... Let's consider an example:
We will write the number 4062 into digits.

4 thousand 0 hundreds 6 tens 2 units or in another way you can write

4062=4 ⋅1000+0 ⋅100+6 ⋅10+2

Next example:
26490=2 ⋅10000+6 ⋅1000+4 ⋅100+9 ⋅10+0

Digit terms are the sum of numbers with different digits.

Let's take the number 86 as an example. Let's split this number into tens and ones. We get: 86 = 80 + 6 = 8 * 10 + 6 * 1. From here we see that the number 86 consists of 8 tens and 6 units. These are the bit terms.

Let's write the division of the bit terms:

• The numbers from 1 to 9 are units;
• Numbers 10, 20,…, 90 are tens;
• The number 100, 200,…, 900 is hundreds and so on.

Any natural number can be divided into bit terms and written as a sum.

Examples of bit terms:

• 892 = 800 + 90 + 2;
• 1695 = 1000 + 600 + 90 + 5;
• 45 = 40 + 5.

Consider an example of determining the bit terms of the number 92586

First, we decompose the number 92586 into bit terms and get:

92 586 = 90000 + 2000 + 500 + 80 + 6 = 9 * 10 000 + 2 * 1 000 + 5 * 100 + 8 * 10 + 6 * 1.

Let's write down what the number 92 586 consists of:

• Out of 9 tens of thousands, 9 * 10,000;
• Out of 2 thousand units 2 * 1000;
• From 5 hundred 5 * 100;
• Out of 8 dozen 8 * 10;
• Of 6 units 6 * 1.

Let us conclude that any number can be divided into bit terms. Bit terms help in solving more complex examples and problems.

A bit term is any natural multi-digit number that can be represented as a sum of bit terms. To decompose a number into bit terms means to divide the number into digits: units, tens, hundreds, thousands, tens of thousands, and so on.

Examples of decomposing numbers into bit terms:

123 = 100 + 20 + 3, where 100 are hundreds, 20 are tens, and 3 are ones.

A more complex example with a large number of digits:

16,458 = 10,000 + 6,000 + 400 + 50 + 8, here 10,000 are tens of thousands, 6,000 are thousands, 400 are hundreds, 50 are tens, 8 are ones.

Number is a mathematical concept for quantitatively describing something or a part of it, it also serves to compare the whole and parts, arrangement in order. The concept of a number is represented by signs or numbers in various combinations. At present, the numbers from 1 to 9 and 0 are used almost everywhere. The numbers in the form of seven Latin letters have almost no use and will not be considered here.

In contact with

Integers

When counting: "one, two, three ... forty-four" or the arrangement in turn: "first, second, third ... forty-fourth" natural numbers are used, which are called natural. This whole set is called "a series of natural numbers" and is denoted by the Latin letter N and does not have an end, because there is always an even larger number, and the largest simply does not exist.

Digits and classes of numbers

Discharges

dozens

• 10…90;

• 100…900.

It can be seen from this that the digit of a number is its position in the digital notation, and any value can be represented through the bit terms in the form nnn = n00 + n0 + n, where n is any digit from 0 to 9.

One ten is the unit of the second category, and one hundred is the third. Units of the first category are called simple, all others are composite.

For the convenience of recording and transmission, the digits are grouped into classes of three in each. It is allowed to put a space between the classes for readability.

Classes

First - units, contains up to 3 characters:

• 200 + 10 +3 = 213.

Two hundred and thirteen contains the following bit terms: two hundred, one ten and three simple ones.

• 40 + 5 = 45;

Forty five consists of four tens and five simple ones.
Second - thousand, 4 to 6 characters:

• 679 812 = 600 000 + 70 000 + 9 000 + 800 +10 + 2.

This sum consists of the following bit terms:

1. six hundred thousand;
2. seventy thousand;
3. nine thousand;
4. eight hundred;
5. ten;

• 3 456 = 3000 + 400 +50 +6.

There are no terms above the fourth category.

Third - million, from 7 to 9 digits:

• 887 213 644;

This number contains nine bit terms:

1. 800 million;
2. 80 million;
3. 7 million;
4. 200 thousand;
5. 10 thousand;
6. 3 thousand;
7. 6 hundred;
8. 4 dozen;
9. 4 units;

• 7 891 234.

In this number, there are no terms above the 7th category.
The fourth is billions, 10 to 12 digits:

• 567 892 234 976;

Five hundred sixty seven billion eight hundred ninety two million two hundred thirty four thousand nine hundred seventy six.

Class 4 bit terms are read from left to right:

1. units of hundreds of billions;
2. tens of billions;
3. units of billions;
4. hundreds of millions;
5. tens of millions;
6. million;
7. hundreds of thousands;
8. tens of thousands;
9. thousand;
10. simple hundreds;
11. simple tens;
12. simple units.

The numbering of the digit of the number is made starting from the lowest, and reading - from the larger.
In the absence of intermediate values ​​in the number of terms, zeros are put in the recording, when pronouncing the name of the missing digits, as well as the class of units, it is not pronounced:

• 400 000 000 004;

Four hundred billion four. The following names of the categories are not pronounced here due to the absence: tenth and eleventh fourth grade; ninth, eighth and seventh third and third grade; also, the names of the second class and its categories, as well as hundreds and tens of units, are not announced.

The fifth is trillions, from 13 to 15 digits.

• 487 789 654 427 241.

Read on the left:

Four hundred eighty seven trillion seven hundred eighty nine billion six hundred fifty four million four hundred twenty seven two hundred forty one.

The sixth is quadrillion, 16-18 digits.

• 321 546 818 492 395 953;

Three hundred twenty one quadrillion five hundred forty six trillion eight hundred eighteen billion four hundred ninety two million three hundred ninety five thousand nine hundred fifty three.

The seventh is quintillion, 19-21 digits.

• 771 642 962 921 398 634 389.

Seven hundred seventy-one quintillion six hundred forty-two quadrillion nine hundred sixty-two trillion nine hundred twenty-one billion three hundred ninety-eight million six hundred thirty-four thousand three hundred eighty nine.

The eighth is sextillion, numbers 22-24.

• 842 527 342 458 752 468 359 173

Eight hundred forty-two sextillions five hundred twenty-seven quintillion three hundred forty-two quadrillion four hundred fifty-eight trillion seven hundred fifty-two billion four hundred sixty-eight million three hundred fifty-nine thousand one hundred seventy-three.

You can simply distinguish between classes by numbering, for example, the number 11 class contains when written from 31 to 33 characters.

But in practice, writing such a number of characters is inconvenient and most often leads to errors. Therefore, when operating with such values, the number of zeros is reduced by raising to a power. After all, it is much easier to write 10 31 than to attribute thirty-one zeros to one.