Upon review students with written multiplication It’s better to take this example for multiplying three- or four digit number to a single digit, where there would be transitions through tens or hundreds, i.e. where oral multiplication is difficult .
Let's take an example: 418 * 3 .
At first students solve it acquaintances them way: replace the first factor sum of bit terms and multiply the sum by the number:
418 * 3 = (400 + 10 + 8) * 3 = 400 * 3 + 10 * 3 + 8 * 3 = 1200 + 30 + 24 = 1254
418 * 3 = (8 + 10 + 400) * 3 = 8 * 3 + 10 * 3 + 400 * 3 = 24 + 30 + 1200 = 1254
After this, the teacher introduces students to written multiplication by single digit number: shows new entry in a column With detailed explanation solutions for the same example.
We need to multiply 418 by 3. We write the second factor under the units of the first factor. We draw a line and put the multiplication sign “X” on the left (it is necessary to explain to the children that multiplication is indicated not only by a dot, but also by such a sign, although a dot can be used here too).
We begin written multiplication with units.
Multiply 8 units by 3 to get 24 units. These are two tens and 4 ones;
We write 4 units under units, and remember 2 tens;
We multiply 1 ten by 3, we get 3 tens, and also 2 tens, we get 5 tens, write them under the tens;
Multiply 4 hundreds by 3 to get 12 hundreds. These are 1 thousand and 2 hundreds.
We write 2 hundreds under hundreds and write 1 thousand in place of thousands.
Work 1254.
From a detailed explanation of the solution to the examples, students, under the guidance of the teacher, move on to a brief explanation when the name of the bit units and the transformations performed is omitted, for example:
578 must be multiplied by 4.
I multiply 8 by 4, it turns out 32. I write 2, and remember 3.
I multiply 7 by 4, it turns out 28, and 3 is only 31; I write 1 and remember 3.
I multiply 5 by 4, it turns out 20, yes 3.
Total 23; I write down 23.
Work 2312.
It can be explained this way: four times eight is thirty-two. 2 I write, 3 I remember.
Four times seven is twenty eight, etc.
You can also write in a line: 578 * 4 = 2312.
At the beginning of studying a topic, the teacher himself informs the students that written multiplication by a single-digit number begins with ones, and later it is useful to explain why written multiplication, like addition and subtraction, begins with the lowest, and not the highest, digit. For this purpose, the same example is solved in two ways:
It turns out that starting written multiplication by a single-digit number with higher-order units is inconvenient, because you have to cross out previously written numbers.
Let's consider cases with zeros in the first factor.
Let's say you need to multiply 42,300 by 6.
The solution to such examples is written as follows:
Explanation:
I sign the second factor 6 under the first non-zero digit of the first factor, under the number 3;
42,300 contains 423 hundreds;
multiply 423 hundreds by 6, we get 2538 hundreds, or 253,800.
When solving similar examples with a detailed explanation, it is necessary to draw children's attention to the fact that in such cases they perform multiplication without paying attention to the zeros written at the end of the first factor, and to the resulting product they add the same number of zeros on the right as they are written at the end of the first factor. At the same time, a brief explanation is given: three times six is 18, I write eight, I remember 1, twice six... I add two zeros to the right, it turns out 253,800.
At this stage, students should also be asked to multiply single-digit numbers by multi-digit numbers: 9 * 136, 4 * 2836, 7 * 1230. When solving such examples, use commutative property of multiplication:
136 * 9, 2836 * 4, 1230 * 7.
Students, having become familiar with written calculation methods, often use them in cases where it is easy to perform calculations orally. It is important to prevent this unwanted transfer. For this purpose, it is necessary to 1) include more relevant cases of multiplication in oral exercises, 2) compare written and oral techniques for multiplying by a single-digit number.
Following multiplication by a single-digit number of natural numbers is the multiplication of quantities expressed in metric units, for example:
9 t 438 kg * 3;
7 km 438 m * 6.
These examples can be solved in different ways: immediately perform the multiplication or first replace the quantities expressed in units of two names with quantities of one name and perform the action:
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9 t 438 kg * 3 = 28 t 314 kg |
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First way more often used in practice when multiplying quantities expressed in units of value
18 rub. 25 kopecks * 3 = 18 rub. * 3 + 25 kopecks. * 3 = 54 rub. 75 kop.
The second method is used when solving problems, as well as in the future when multiplying quantities by any two-digit or three-digit number.
Methodology for studying a written multiplication algorithm (stage 2).
II stage. Multiplication by digit numbers .
After students have a firm grasp of single-digit multiplication, techniques for multiplying by 10, 100, 1000, and then 40, 400, and 4000 are covered.
When multiplying by two- to four-digit place numbers, use property of multiplying a number by a product, For example:
14 * 60 = 14 * (6 * 10) = 14 * 6 * 10 = 840.
To become familiar with this property, students are asked to calculate in different ways the value of the expression is 16 * (5 * 2). Under the guidance of a teacher, they find the meaning of an expression in these ways;
16 * (5 * 2) = 16 * 10 = 160
16 * (5 * 2) = (16 * 5) * 2 = 80 * 2 = 160
16 * (5 * 2) = (16 * 2) * 5 = 32 * 5 = 160
Students notice that
in the first case, they multiplied the number 16 by the product of the numbers 5 and 2;
in the second, the number 16 was multiplied by the first factor 5 and the resulting product was multiplied by the second factor 2;
in the third - the number was multiplied by the second factor 2 and the resulting product was multiplied by the first factor 5;
the meanings of the expressions are the same.
After completing several such exercises, students formulate the property: “To multiply a number by a product, you can find the product and multiply the number by the result obtained, or you can multiply the number by one of the factors and multiply the result by another factor.”.
The property of multiplying a number by a product is used when performing various exercises:
solving examples and problems in various ways, for example:
in a convenient way, for example: 25 * (2 * 7) = (25 * 2) * 7 = 350;
comparison of expressions, for example. 24 * 5 * 10 and 24 * 50, etc.
This property is then used to disclosure of the computational method of multiplication into two-digit and four-digit digit numbers.
Preparatory exercises are first introduced to replace digit numbers with the product of a single-digit number and 10 (100, 1000), for example: 70 = 7 * 10, 600 = 6 * 100.
Next, oral techniques for multiplying by place numbers are discussed. For example, you need to multiply 15 by 30; Let's imagine the number 30 as a product of convenient factors 3 and 10, we get an example: 15 multiplied by the product of the numbers 3 and 10; here it is more convenient to multiply the number 15 by the first factor - by 3 and the resulting result 45 multiplied by the second factor - by 10, you get 450. Entry:
15 * 30 = 15 * (3 * 10) = (15 * 3) * 10 = 450
Students sometimes mix the property of multiplying a number by a product with the property of multiplying a number by a sum.
For example, an error of the form 15 * 12 = 300 indicates such a confusion: the student multiplies 15 by 2 and multiplies the resulting result by 10, i.e. he replaced the number 12 with the sum of the bit terms 10 and 2, and then multiplied by both the product of these numbers, i.e. to the number 20.
A similar error also occurs when performing exercises to compare expressions, for example:
27 * 7 * 10 = 27 * 7 + 27 * 10
To prevent such errors, it is useful to offer exercises to compare relevant calculation techniques. For example, students solve the following examples with commentary and detailed recording:
6 * 50 = 6 * (5 * 10) = 6 * 5 * 10 = 300
6 * 15 = 6 * (10 + 5) = 6 * 10 + 6 * 5 = 90
Then it turns out that both examples have the same first factors, but different second ones; when solving examples, the second factor (50) was replaced by the product of convenient factors (5 and 10) and the property of multiplying a number by a product was used: the number 6 was multiplied by the first factor and the resulting product was multiplied by the second factor. In the second example, the factor 15 was replaced by the sum of the digit terms 10 and 5 and the property of multiplying a number by a sum was used; multiplied the number 6 by the first term, then multiplied the same number 6 by the second term and added the results.
It is also useful to offer children exercises to compare expressions (put “>” instead of empty cells, “<» или « = »):
36 * 10 * 4 □ 36 * 14 17 * 5 * 10 □ 17 * 50
45 * 6 + 45 * 10 □ 45 * 60 16 * 10 □ 16 * 3 +16 * 10
21 * 4 + 21 * 3 □ 21 * 12 18 * 9 + 18 * 10 □ 18 * 19
In order to prevent errors in mixing the properties of arithmetic operations studied in elementary grades, it is necessary to perform exercises comparing them more often.
After learning the techniques of oral multiplication by place numbers, the techniques of written multiplication are introduced. It is proposed to solve the example 546 * 30.
Let's calculate in writing, write the example like this:
First multiply the number 546 by 3, and multiply the resulting result by 10. Multiply 546 by 3:
three times six - 18; eight we write, 1 we remember;
three times four - 12, yes 1, it turns out 13, write three, remember 1;
three times five is 15, yes 1, it turns out 16, write 16, we get 1638.
We multiply 1638 by 10, to do this we add one zero to the right of the resulting number.
Product 16 380.
Note that here, when multiplying by a single-digit number (546 * 3), we use a brief explanation. The same should be done in the future, when in new, more complex cases of multiplication, multiplication by a single-digit number is an integral part.
Multiplying by three- and four-digit digits works in the same way as multiplying by two-digit digits.
Particularly noteworthy are those cases in which both factors end in zeros, for example: 20 30, 400 50, 800 70, 4000 60, etc.
First, when solving such examples, students reason as follows: to multiply 300 by 50, you need to multiply 3 hundreds by 5, and then multiply the resulting number by 10, which will be 150 hundreds, or 15,000.
Such examples are written down on a line and solved orally.
Students reason in a similar way when doing written multiplication in the case when both factors end in zeros.
It is more convenient to write such examples in a column as follows:
Observing the multiplication of numbers ending in zeros, students come to the conclusion that first in these cases it is necessary to multiply the numbers that will be obtained if these zeros are discarded, and then to the resulting product add as many zeros on the right as are written at the end of both factors together. In the future, when multiplying numbers ending in zeros, students are guided by this conclusion.
Methodology for studying a written multiplication algorithm (stage 3).
Lesson (ONZ)
find values of numeric expressions
fulfill division with remainder;
fulfill tasks of a searching and creative nature
The main structural elements of the lesson:
New knowledge: algorithm for dividing by a single-digit number by an angle.
Trial action: find the quotient when writing division by angle (No. 3 (a) (RT), p. 12).
Fixing a problem: “I still can’t find the quotient of the angle.”
Fixing the cause of the difficulty: “I don’t know how to divide by an angle.”
Purpose of student activity: “Learn how to divide by angle.”
Recording new knowledge : in speech and with the help standard in which the angle division algorithm is fixed.
In this lesson, we are training the skills of constructing a speech, a statement, taking into account
Topic: “Multiplying by a single digit number”
Main goals:
Personal:
1) To help students understand the practical value of the knowledge acquired.
Metasubject:
1) Develop the ability to use the simplest techniques of oratory.
2) Train the ability to conduct observation in educational activities.
Subject:
1) Develop the ability to multiply by a single-digit number.
Lesson progress:
1. Motivation for learning activities.
Lesson motto on the board
Let us proceed to multiplication with particular importance,
Let's share the joy of discovery with a friend.
Read the motto. What will the lesson be about? ( We will continue to work with the action of multiplication.)
What discovery are we talking about? (On the discovery of new knowledge.)
Why does the motto talk about the joy of discovery? (Because discovering new things brings joy.)
2. Updating knowledge and fixing individual difficulties in a trial action.
What should you do before learning something new? (Repeat what we know.)
Will we repeat everything we know? (No, only what will help us in discovering new knowledge.)
To update knowledge, tasks are completed.
The tasks are completed by students; in pairs, the test is carried out frontally with justification.
Solving tasks:
Divide the numbers into groups:
(single-digit and double-digit, even, odd)
(6+7)*3 1dec.4 units=
(10+2)*2= 3dec.7 units=
What did you repeat? ( ...)
What task awaits you ahead? (Task for a trial action.)
Test task
Do 21*4
Who didn't make it? Who doesn't have an answer?
What is your problem? (We cannot yet multiply a two-digit number by a one-digit number.)
Who received the answer, what rule did you use?
What is your problem? (I cannot name the standard I used.)
3. Identifying the location and cause of the difficulty.
What should be done now? (We need to think about it.)
What task were you doing? (multiplying by a single digit number.)
Why did the problem arise? ((We don't know how to multiply by a single digit number.)
4. Construction of a project for getting out of the difficulty.
What goal will you set for yourself now? ((Open a method for multiplying by a single-digit number.)
The teacher writes the purpose of the lesson on the board .
Formulate the topic of the lesson. ( Multiplying by a single digit number)
What will this method help you discover? (Knowledge of multiplication tables, multiplication by 10)
An example of performing multiplication will also help you, which can be seen in ... (textbook) p. 55
What should you do? (analyze the sample, see how the Hare and the Wolf performed the multiplication. read the steps of the algorithm and compose the algorithm, draw a conclusion.)
5. Implementation of the constructed project.
So, the first point of the plan is to analyze the sample.
In what order do you think the calculations were carried out, what was done first, what then?
Well done, now you can simulate the new algorithm.
By groups:
Represent two-digit values by sum of tens and ones
.Multiply each term by a number
Add up the results
The teacher records the note on the board.
Can you solve similar examples this way? (Yes.)
Let's write down the expression of the trial action in a workbook with comments on the algorithm.
21*4=(20+1)*4=20*4+1*4=80+4=84
Have you overcome the difficulty? (Yes.)
How is multiplication written differently?
(in a column)
Consider the column entry from top to bottom. In what sequence do you think the calculations were carried out, what digit units were multiplied first. what then?,
Now you can simulate the new algorithm
.Units are multiplied first, then tens.
6. Primary consolidation in external speech.
For initial consolidation, you can suggest doing №2
Column, p. 56 by reciting the steps of the algorithm.
In pairs No. 2 (2nd column). Self-test is carried out step by step according to the sample
Who completed the task correctly? Put a "+" sign. Who had any mistakes? Put "?". Did you understand the cause of the error and correct it? You also completed the task, put the “+” sign.
7. Independent work with self-test according to the standard.
You did a very good job. What should you do now? (Do independent work.)
For what? (To test yourself.)
Independent work: Notebook, No. 85 page 25
After completing the independent work, self-test according to the detailed self-test sample
Analysis and correction of errors.
Those students who did not make mistakes put a “+” sign, those who made mistakes put a “?+” sign.
8. Inclusion in the knowledge system and repetition.
Where in math lessons do you encounter multiplying a two-digit number by a one-digit number? (In solving problems.)
Now let's turn our attention to the tasks. Read the problem statement No. 8 page 57 of the textbook What method will be used to solve this problem? (Algorithm for multiplying by a single digit number
The problem can be proposed for solution individually
Students check their answer using the model.
Solution of the task.
1)6*3=18(dm)-length
2)18*6=(10+8)*6-60+48=108 sq.dm
Answer: sheet area 108 sq. dm
9. Reflection on activities in the lesson.
What was your goal for the lesson? (Discover a method for multiplying by a single-digit number.)
Have you achieved your goal? (Yes.)
Prove it. (Completed independent work.)
How did you discover new knowledge? (We performed a trial action, realized that we do not know, discovered new knowledge.)
5b - I didn’t understand the topic. there are errors
6b - I understand the topic, but there are errors
7b - understood the topic, no errors, but did not remember the algorithm
8b - understood the topic, no errors, remembered the algorithm
Homework (Slide 15 ):
Required part: remember the multiplication algorithm,
No. 86(RT), p. 25,
2.Optional tasks:
Lecture 3. Algorithms for arithmetic operations on non-negative integers in the decimal number system.
1. Addition algorithm
2. Subtraction algorithm
3. Multiplication algorithm
4. Division algorithm
An algorithm is one of the fundamental concepts that is used in various fields of knowledge, but it is studied in mathematics and computer science. In our course we will use an intuitive and meaningful interpretation of the concept of “algorithm”, according to which we will consider an algorithm as a program of action for solving problems of a certain type.
Mastering the algorithm begins already in elementary school during mathematics lessons, where students master algorithms for arithmetic operations, become familiar with the rules for subtracting a number from a sum, a sum from a number, etc.
In general, the formation of algorithmic thinking in younger schoolchildren is currently one of the most important tasks of a teacher, and therefore he requires certain knowledge about algorithms, as well as some skills in their construction.
Addition algorithm
The addition of single-digit numbers can be performed based on the definition of this action, but in order not to have to resort to the definition every time, all the sums that are obtained by adding single-digit numbers are recorded in a special table called single digit addition table, and remember.
Naturally, the meaning of addition remains the same for multi-digit numbers, but the practical implementation of addition occurs according to special rules. The sum of multi-digit numbers is usually found by performing column addition.
Example 7.
Let us find out how this algorithm arises and what theoretical principles underlie it.
Let's imagine the terms 341 and 7238 as a sum of powers of ten with coefficients:
341+ 7238 = (3∙10 2 + 4∙10 + 1) + (7∙10 3 + 2∙10 2 +3∙10 + 8).
Let's open the parentheses in the resulting expression, swap places and group the terms so that the units are next to the ones, tens are next to tens, etc. All these transformations can be performed based on the corresponding properties of addition. Associativity property allows you to write an expression without parentheses:
3∙10 2 + 4∙10 + 1 + 7∙10 3 + 2∙10 2 + 3∙10+8.
Based on commutative properties Let's swap the terms:
7∙10 3 + 3∙10 2 + 2∙10 2 + 4∙10+3∙10 + 1 + 8.
According to associativity property Let's make a grouping:
7∙10 3 + (3∙10 2 + 2∙10 2) + (4∙10 + 3∙10) +(1 + 8).
Let's take the number 10 2 out of brackets in the first selected group, and 10 in the second. This can be done in accordance with property of distributivity of multiplication relative to addition:
7∙10 3 + (3 + 2)∙10 2 + (4 + 3)∙10 + (1 + 8).
So, the addition of these numbers 341 and 7238 was reduced to the addition of single-digit numbers represented by the digits of the corresponding digits. We find these amounts using the addition table:
7∙10 3 +5∙10 2 + 7∙10 + 9.
The resulting expression is the decimal notation of the number 7579.
We see that The algorithm for adding multi-digit numbers is based on the following theoretical facts :
- a way to write numbers in the decimal number system;
- properties of commutativity and associativity of addition;
- distributivity of multiplication relative to addition;
- table of addition of single-digit numbers.
It is easy to verify that in the case of adding numbers “with passing through ten” the theoretical foundations of the addition algorithm will be the same.
In general algorithm for adding natural numbers written in the decimal number system, formulated as follows:
1) Write the second term under the first so that the corresponding digits are located under each other.
2) Add the units of the first digit. If the amount is less than ten, write it down in the answer units category and move on to the next category (tens).
3) If the sum of units is greater than or equal to ten, then it is represented as a 0 + b 0 = 1∙10 + c 0, where c 0 is a single-digit number; write with 0 in the units category of the answer and add 1 to the tens of the first term, after which they move to the tens category.
4) Repeat the same steps with tens, then with hundreds, etc. The process ends when the most significant digits are added. Moreover, if their sum is greater than or equal to ten, then we assign zeros in front of both terms, increase the zero in front of the first term by 1 and perform the addition 1+0=1.
Note that in this algorithm (as in some others) for brevity, the term “digit” is used instead of “a single-digit number represented by a digit.”
Subtraction algorithm
Example 8. Let's consider the difference between the numbers 485 and 231.
Let's use the rule for writing numbers in the decimal number system and present this difference in this form:
485-231= (4∙10 2 +8∙10+5)-(2∙10 2 +3∙10+1). To subtract the sum 2∙10 2 +3∙10+1 from the number 4∙10 2 +8∙10+5, it is enough to subtract from it each term of this sum one by one, and then:
(4∙10 2 +8∙10+5)-(2∙10 2 +3∙10+1)=(4∙10 2 +8∙10+5)-2∙10 2 -3∙10-1.
To subtract a number from a sum, it is enough to subtract it from any one term(greater than or equal to this number). Therefore, we subtract the number 2∙10 2 from the term 4∙10 2, the number 3∙10 from the term 8∙10, and the number 1 from the term 5, then:
(4∙10 2 +8∙10+5)-2∙10 2 -3∙10-1=(4∙10 2 -2∙10 2)+(8∙10-3∙10)+(5-1).
Let's take advantage and put 10 2 and 10 out of brackets. Then the expression will look like:
(4-2) ∙10 2 +(8-3) ∙10+(5-1).
We see that the subtraction of the three-digit number 231 from the three-digit number 485 has been reduced to the subtraction of single-digit numbers represented by the digits of the corresponding digits in the notation of the given three-digit numbers. We find the differences 4-2, 8-3 and 5-1 using the addition table and obtain the expression: 2∙10 2 +5∙10+4, which is a representation of the number 254 in the decimal number system. Thus, 485-231=254.
The expression (4-2)∙10 2 +(8-3)∙10+(5-1) specifies the subtraction rule, which is usually carried out by a column:
We see that subtraction multi-digit number from polysemantic based on :
- the method of writing numbers in the decimal number system;
- rules for subtracting a number from a sum and a sum from a number;
- the distributive property of multiplication relative to subtraction;
- single-digit addition table.
It is not difficult to verify that if in some place of the minuend there is a single-digit number less than a number in the same place of the subtrahend, then the subtraction is based on the same theoretical facts and the table for adding single-digit numbers.
Example 9. Let's find the difference between the numbers 760-326.
Let's use the rule for writing numbers in the decimal number system and present this difference in this form:
760-326= (7∙10 2 +6∙10+0)-(3∙10 2 +2∙10+6).
Since 6 cannot be subtracted from the number 0, it is impossible to perform a subtraction similar to what was done in the first case. Therefore, let’s take one ten from the number 760 and present it as 10 units (the decimal number system allows us to do this). Then we will have the expression:
(7∙10 2 +5∙10+10)-(3∙10 2 +2∙10+6).
If we now use rules for subtracting an amount from a number And numbers from the sum, and also distributivity of multiplication relative to subtraction, then we get the expression (7-3) ∙10 2 +(5-2) ∙10+(10-6) or 4∙10 2 +3∙10+4.
The last sum is the number 434 written in decimal notation. So, 760-326=434.
Algorithm for subtracting numbers in the decimal number system:
1. We write the subtrahend under the minuend so that the corresponding digits are under each other.
2. If the digit in the units digit of the subtrahend does not exceed the corresponding digit of the minuend, subtract it from the digit of the minuend, write the difference in the units digit of the desired number, and then move on to the next digit.
3. If the number of the subtrahend is greater than the units of the minuend, i.e. b 0 >a 0, and the tens digit of the minuend is different from zero, then we reduce the tens digit of the minuend by 1, while simultaneously increasing the units digit of the minuend by 10, after which we subtract from the number 10+ a 0 number b 0 and write down the difference in the units digit of the desired number, then move on to the next digit.
4. If the number of units of the subtrahend is greater than the number of units of the minuend, standing in the tens, hundreds, etc. place. minuend are equal to zero, then we take the first non-zero digit in the minuend (after the units place), reduce it by 1. All digits in the lower digits up to and including the tens place are increased by 9, and the digit in the units place by 10: subtract b 0 out of 10+ a 0, write down the difference in the units digit of the desired number and move on to the next digit.
5. In the next category we repeat the described process.
6. Subtraction ends when the minuend is subtracted from the most significant digit.
Multiplication algorithm
Multiplication of single digit numbers can be done based on the definition of this operation. But in order not to have to resort to the definition every time, all products of single-digit numbers are written in a special table called multiplication table single-digit numbers, and remember.
Naturally, the meaning of multiplication remains the same for multi-digit numbers, but the calculation technique changes. The product of multi-digit numbers is usually found by performing column multiplication using a specific algorithm. Let's find out how this algorithm arises and what theoretical facts underlie it.
Example 10.
Let's multiply, for example, with a column of 428 by 263. 428
We see that to get the answer we had to x 263
Multiply 428 by 3.6, and 2, i.e. multiply multi-digit 1284
number to single digit; but multiplied by 6, the result is +2568
written in a special way, placing the units of the number 856__
2568 under the tens of the number 1284, since we multiplied 112564
by 60 and got the number 25680, but omitted the zero at the end of the entry.
The term 856 is the result of multiplication by 2 hundreds, i.e. number 85600.
In addition, we had to find the sum of multi-digit numbers.
So, in order to multiply a multi-digit number by a multi-digit number, you must be able to:
Multiply a multi-digit number by a single-digit number and by a power of ten;
Add multi-digit numbers.
First, let's look at multiplying a multi-digit number by a single-digit number.
Example 11.Multiply 428 by 3.
According to the rule for writing numbers in the decimal number system 428 can be represented as
4∙10 2 +2∙10+8 and then 428∙3=(4∙10 2 +2∙10+8) ∙3.
Based on distributivity of multiplication relative to addition Let's open the brackets:
(4∙10 2) ∙3+(2∙10) ∙3+8∙3.
Products in brackets can be found by multiplication table single-digit numbers: 12∙10 2 +6∙10+24. We see that multiplying a multi-digit number by a single-digit number has been reduced to multiplying single-digit numbers. But to get the final result, you need to transform the expression 12∙10 2 +6∙10+24 - the coefficients in front of the powers of 10 must be less than 10.
To do this, imagine the number 12 as 1∙10+2, and the number 24 as 2∙10+4. Then in the expression (1∙10+2)∙10 2 +6∙10+(2∙10+4) we open the brackets: 1∙10 3 +2∙10 2 +6∙10 +2∙10+4. Based on the associativity of addition and the distributivity of multiplication relative to addition Let's group the terms 6∙10 and 2∙10 and put 10 out of brackets: 1∙10 3 +2∙10 2 +(6+2) ∙10+4.
The sum 6+2 is the sum of single-digit numbers and can be found using the addition table: 1∙10 3 +2∙10 2 +8∙10+4. The resulting expression is the decimal notation of the number 1284, i.e. 428∙3=1284.
Thus, Multiplying a multi-digit number by a single-digit number is based on:
- recording numbers in the decimal number system;
- properties of addition and multiplication;
- tables of addition and multiplication of single-digit numbers.
Multi-digit multiplication algorithm a n a n -1 …a 1 a 0 to a single digit number at .
1. Write the second number under the first.
2. Multiply the digits of the units place of a number X. per number u. If the product is less than 10, we write it in the units category of the answer and move on to the next category (tens).
3. If the product of digits of units of a number X. per number at is greater than or equal to 10, then we represent it in the form 10q 1 +c 0, where c 0 is a single-digit number; We write from 0 to the response units category and remember q 1 - transfer to the next category.
4. Multiply the tens digits by the number at, add the number q 1 to the resulting product and repeat the process described in paragraphs 2 and 3.
5. The multiplication process ends when the highest digit is multiplied.
Let's consider an algorithm for multiplying a multi-digit number by a multi-digit number.
Example 12. Let's calculate the product 428∙263.
Let's imagine the number 263 as the sum 2∙10 2 +6∙10+3 and write the product 428∙ (2∙10 2 +6∙10+3). It, according to the distributivity of multiplication relative to addition, equals 428∙(2∙10 2)+428∙(6∙10)+428∙3.
From here, applying associative property of multiplication, we get:
(428∙2) ∙10 2 +(428∙6) ∙10+428∙3.
We see that multiplying the multi-digit number 428 by the multi-digit number 263 has been reduced to multiplying the multi-digit number 428 by the single-digit numbers 2,6 and 3, as well as by powers of 10.
Algorithm for multiplying a multi-digit number by a multi-digit number:
1. Write down the multiplier X and below it the second factor at.
2. Multiply the number X to the junior rank b 0 numbers at and record the piece x*b 0 under the number at.
3. Multiply the number X to the next level b 1 numbers at and record the piece xb 1, but shifted one bit to the left, which corresponds to multiplication xb 1 by 10.
4. We continue calculating the products until calculating xb k . .
5. We add the resulting k+1 products.
Studying the algorithm for multiplying multi-digit numbers in an initial mathematics course, as a rule, takes place in accordance with the identified stages. The only differences are in the recording. For example, when justifying the case of multiplying a multi-digit number by a single-digit number, they write: 428∙3=(400+20+8) ∙ 3=400∙3+20∙3+8∙3=1200+60+24=1284.
The basis of the transformations performed are:
- representation of the first multiplier as a sum of digit terms (i.e., writing a number in the decimal number system);
- the rule for multiplying a sum by a number (or the distributivity of multiplication relative to addition);
- multiplication of “round” (i.e. ending in zeros) numbers by a single-digit number, which comes down to multiplying single-digit numbers.
Division algorithm
When it comes to the technique of dividing numbers, this process is considered as action of division with remainder : divide a non-negative integer A to a natural number b– this means finding such non-negative integers q And r, What a=bq+r, and 0< r .
