If you have already forgotten how to multiply numbers in a column, then read the article. Here you will find all the information about this mathematical operation.

Even some adults have not mastered in school how to multiply numbers in a column. But this skill can come in handy in life if you don’t have a calculator or a mobile phone at hand.

Moreover, this is not at all difficult if you know the multiplication table and understand how to correctly arrange the numbers in this process. Multiplication in a column always begins to be studied by multiplying a multi-digit number by a single-digit number in order to understand the rules of this action. Further details.

Rules and algorithm for multiplication in a column

Mathematical classes for many children are not given the first time. This is a complex science that requires special attention and understanding. And students in the primary grades without fail need the help of mom and dad in solving complex examples and tasks. In particular, you cannot leave everything to chance if your child does not understand what multiplication, division of numbers, etc. are. It is necessary to help understand the topic and learn the multiplication table, so that later you do not get bad grades, and do not get upset.

It will be easy to master multiplication in a column if:

• The student knows the multiplication table very well. Do not get confused in the meanings of the work.
• I figured out in what sequence the digits of a multi-digit number should be multiplied.
• The child understood where to write them correctly. And he knows how to add polynomials in a column.

You need to know the rule that the product does not change from a change in the places of factors. More precisely, if you multiply 56 ⋅ 2 = 112 and 2 ⋅ 56 = 112, the product will be 112.

IMPORTANT: When multiplying numbers in a column. Below the bottom they write the number that has fewer digits in its composition.

How to correctly multiply three-digit numbers by one-digit, two-digit, three-digit numbers in a column

Any multiplication is the addition of identical numbers the necessary number of times. More precisely, 725 ⋅ 2 \u003d 725 + 725 \u003d 1450. But such an example can be done verbally if the second number is 2,3,4. And if it is 8, then it is already better to multiply in a column. For this:

1. At the top you need to write a number 725 , and below under the number - 5 write the number - 8.
2. Now you need to take turns starting from 5, all values ​​of a three-digit number multiply by 8.
3. More precisely: 5 ⋅ 8 = 40 ( write zero below under the eight and five, and remember 4).
4. Then we multiply: 2 ⋅ 8 = 16 ( to 16 we add - 4 \u003d 20, again we write 0, only under the deuce, and - 2 we remember).
5. It remains to multiply: 7 ⋅ 8 = 56 ( add to 56 - 2 \u003d 58, write the eight under the seven, and five in front).
6. As a result of this multiplication ( 725 ⋅ 8 ) get - 5800 . And this calculation was obtained manually, without any machines, calculators.

Multiplication in a column - three-digit by three-digit

Multiplying a polynomial by a polynomial is somewhat more difficult. However, if you have already figured out how the process takes place in the first example, then it will not be difficult for you to multiply three-digit numbers, and then add the resulting values ​​​​in a column.

Consider in detail how to multiply 125 by 32

1. At the top of the sheet, write the three-digit number 125, below it 32, and arrange it as follows: three under the two of the first number, A two of the second under the five of the first number- it is very important.
2. Start multiplying from the end. That is: multiply all digits of a three-digit number(125) first on deuce.
3. You get 250, zero write under the deuce, the rest of the numbers are ahead.
4. Further multiply 125 by three. And place on the leaf the value of the work ( 375 ), starting with the number - 3 .
5. Now it remains to fold 250 and 375(0), it will turn out 250 + 3750 = 4000.

IMPORTANT: How to multiply three-digit numbers can be clearly seen in the figure above. The numbers are multiplied in strict sequence, starting from the end, and then all the resulting values ​​\u200b\u200bare added up.

How to correctly multiply numbers with zeros in a column?

Already from elementary school mathematics, any student knows that if you multiply any number by zero, then the product will also be 0. That is why, when multiplication is performed in a column, then multiplication is not performed by the number zero, it is taken out of the frame, and in the work it is attributed zero or more zeros - see the image below.

How to explain to a child multiplication by a column?

• If you decide to have a math lesson at home, learn how to multiply in a column, then turn your lesson into a game.
• Gradually, patiently explaining how it's done. Answer all the questions of the student so that he understands what and why to do.
• First, give simple examples for examples, and then choose tasks that are more difficult.

IMPORTANT: Spend more time with your children, do not ignore their requests for help. At school, the teacher complies with the program requirements. Not much time is given to consolidate the material. Therefore, not all students have time to master the program, especially in such a complex matter as multiplication, division in a column.

Video: Examples of multiplying multi-digit numbers in a column with explanations

Don't like math? You just don't know how to use it! In fact, it is a fascinating science. And our selection of unusual multiplication methods confirms this.

Multiply on your fingers like a merchant

This method allows you to multiply numbers from 6 to 9. First, bend both hands into fists. Then, on the left hand, bend as many fingers as the first factor is greater than the number 5. On the right, do the same for the second factor. Count the number of extended fingers and multiply the amount by ten. Now multiply the sum of the bent fingers of the left and right hands. Adding both sums, you get the result.

Example. Multiply 6 by 7. Six is ​​more than five by one, which means we bend one finger on the left hand. And seven - two, so on the right - two fingers. In total, this is three, and after multiplying by 10 - 30. Now we multiply four bent fingers of the left hand and three - of the right. We get 12. The sum of 30 and 12 will give 42.

In fact, here we are talking about a simple multiplication table, which would be nice to know by heart. But this method is good for self-examination, and stretching your fingers is useful.

Multiply like Ferrol

This method was named after the German engineer who used it. Method allows you to quickly multiply numbers from 10 to 20. If you practice, you can do it even in your mind.

The point is simple. The result will always be a three-digit number. So first we count the ones, then the tens, then the hundreds.

Example. Multiply 17 by 16. To get units, we multiply 7 by 6, tens - we add the product of 1 and 6 with the product of 7 and 1, hundreds - we multiply 1 by 1. As a result, we get 42, 13 and 1. For convenience, we write them in a column and add up. Here is the result!

Multiply like a Japanese

This graphic method used by Japanese schoolchildren allows you to easily multiply two- and even three-digit numbers. Get some paper and a pen ready to try it out.

Example. Multiply 32 by 143. To do this, draw a grid: reflect the first number with three and two horizontally indented lines, and the second with one, four and three vertically indented lines. Place dots where the lines intersect. As a result, we should get a four-digit number, so we will conditionally divide the table into 4 sectors. And recalculate the points that fall into each of them. We get 3, 14, 17 and 6. To get the answer, add the extra ones for 14 and 17 to the previous number. We get 4, 5 and 76 - 4576.

Multiply like an Italian

Another interesting graphic method is used in Italy. Perhaps it is simpler than Japanese: you definitely won’t get confused when transferring dozens. To multiply large numbers with it, you need to draw a grid. We write the first multiplier horizontally from above, and the second one vertically to the right. In this case, there should be one cell for each digit.

Now multiply the numbers in each row by the numbers in each column. We write the result in a cell (divided in two) at their intersection. If you get a single-digit number, then write 0 in the upper part of the cell, and the result obtained in the lower part.

It remains to add up all the numbers that are in the diagonal stripes. We start from the bottom right cell. At the same time, tens are added to the units in the next column.

Here's how we multiplied 639 by 12.

Fun, right? Have fun with mathematics! And remember that the humanities in IT are also needed!

It is convenient to multiply multi-digit or multi-digit numbers in writing in a column, multiplying each digit in succession. Let's see how to do it. Let's start by multiplying a multi-digit number by a single-digit number and gradually increase the capacity of the second multiplier.

To multiply two numbers in a column, place them one below the other, ones under ones, tens under tens, and so on. Compare two factors and place the smaller one under the larger one. Then start multiplying each bit of the second multiplier by all the bits of the first multiplier.

Multiplication of a multi-digit number by a single-digit number

We write a one-digit number under the units of a multi-digit number.

Multiply 2 sequentially to all digits of the first multiplier:

Multiply by units:

8 x 2 = 16

6 write under units, and 1 remember ten. In order not to forget, we write 1 over dozens.

Multiply by tens:

3 tens × 2 = 6 tens + 1 tens (remembered) = 7 tens. We write the answer under tens.

Multiply by hundreds:

4 hundreds × 2 = 8 hundreds . We write the answer under hundreds. As a result, we get:

438 x 2 = 876

Multiplication of a multi-digit number by a multi-digit number

Multiply a three-digit number by a two-digit number:

924×35

We write a two-digit number under a three-digit one, units under units, tens under tens.

Stage 1: find the first incomplete product, multiplying 924 on 5 .

Multiply 5 sequentially to all digits of the first multiplier.

Multiply by units:

4 x 5 = 20 0 we write under the units of the second multiplier, 2 remember ten.

Multiply by tens:

2 tens × 5 = 10 tens + 2 tens (remembered) = 12 tens , we write 2 under the tens of the second multiplier, 1 remember.

Multiply by hundreds:

9 hundreds × 5 = 45 hundreds + 1 hundred (remembered) = 46 hundreds, we write 6 under the hundreds digit, and 4 under the thousands place of the second multiplier.

924 × 5 = 4620

Stage 2: find the second incomplete product, multiplying 924 on 3 .

Multiply 3 sequentially to all digits of the first multiplier. We write the answer under the answer of the first stage, shifting it one place to the left.

Multiply by units:

4 x 3 = 12 2 write under the tens place, 1 remember.

Multiply by tens:

2 tens × 3 = 6 tens + 1 tens (remembered) = 7 tens, we write 7 under the hundreds digit.

Multiply by hundreds:

9 hundreds × 3 = 27 hundreds , 7 write in the thousands place, and 2 into the tens of thousands.

Stage 3: add both incomplete products.

We add bit by bit, taking into account the shift.

As a result, we get:

924 × 35 = 32340

Multiply a three-digit number by a three-digit number:

Let's take the first factor from the previous example, and the second factor from the previous one, but 8 hundred more:

924×835

So, the first two steps are the same as in the previous example.

Stage 3: find the third incomplete product, multiplying 924 on 8

Multiply 8 sequentially to all digits of the first multiplier. We write the result under the second incomplete product shifted to the left, to the hundreds place.

4 x 8 = 32, we write 2 into the hundreds 3 remember

2 x 8 = 16 + 3(remembered) = 19 , we write 9 in the ranks of thousands 1 remember

9 x 8 = 72 + 1(remembered) = 73 , we write 73 into the hundreds and tens of thousands, respectively.

Stage 4: add three incomplete products.

As a result, we get:

924 × 835 = 771540

So, how many digits are in the second factor, there will be so many terms in the sum of incomplete products.

Let's take two multipliers with the same bit depth:

3420×2700

When multiplying two numbers ending in zeros, we write one number under the other so that the zeros of both factors are left out.

Now we multiply two numbers, ignoring the zeros:

342 × 27 = 9234

We attribute the total number of zeros to the resulting product.

As a result, we get:

3420 × 2700 = 9234000

Summarize. In order to multiply two numbers in a column in writing, you need to :

1. Compare two numbers and write the smaller one under the larger one, units under units, tens under tens, and so on. If there are numbers with zeros, then we write one number under the other so that the zeros of both factors are left out.

2. We multiply successively each bit of the second factor, starting from units, by all the bits of the first multiplier. We don't pay attention to zeros.

3. We write incomplete works one under the other, shifting each incomplete work one digit to the left. How many significant digits (not 0) are in the second multiplier, so many incomplete products will be.

4 . We add up all the incomplete works.

5. We assign zeros from both factors to the result obtained.

That's all, thanks for being with us!

Let's write the numbers in a column (one below the other). The top row is the larger number, the bottom row is the smaller number.

The rightmost digit (sign) of the top number must be above the rightmost digit of the bottom number. On the left side, between the numbers, we put the sign of the action. We have this "×" (multiplication sign).
First, multiply the entire top number by the last digit of the bottom number. The result is written under the line under the rightmost digit.

Multiply the number from above by the digit (sign) from right to left.

We got a number greater than or equal to "10".

Therefore, only the last digit of the result goes below the line. This is "2". The number of tens of the product (we have “4 tens”) is placed above the neighbor to the left of “7”.
We multiply "2" by "6".

The result of multiplication by the second digit must be written under the second digit of the result of the first multiplication.

Now having mastered multiplication by a column, you can multiply arbitrarily large numbers.

MULTIPLICATION BY A COLUMN OF TWO-DIGITAL NUMBERS

Math simulator

The program is a simulator in mathematics to consolidate skills multiplication by a column of two-digit numbers.

There are 20 examples to solve. Two random two-digit numbers need to be multiplied by a column.

To go to the beginning of solving examples, press the "START" button

At the top left of the math simulator page is the number of examples left to solve.

On the right side of the page is an example to solve. On the left side, the same example is written in a column.

Use the cursor keys to move up/down/right/left through the cells. Press the buttons 0-9 on the keyboard and enter the intermediate answers and the final answer.

If the example is solved correctly, 5 points are awarded. If you give the correct answer three times in a row, a bonus is awarded.

3 points are deducted for an incorrect answer.

Errors made during the calculation are corrected in red. It will be immediately clear at what stage of the calculations an error was made.

The final page of the simulator in mathematics presents the results: the number of points, errors, bonuses.

If at column multiplication errors were made, examples in which they were will be listed below.

Rules for multiplying two-digit numbers by a column

Method column multiplication, allows you to simplify the multiplication of numbers. Multiplication by a column suggests sequential multiplication of the first number, to all digits of the second number of the subsequent addition of the resulting products, taking into account indentation, depending on the position of the digit of the second number.

Consider how to multiply by a column using the example of finding the product of two numbers 625 × 25 .

With more digits in the second number, we get that our works line up on the right in the form of a "ladder".

4 As a result of multiplication, we get 2 works, 3125 And 1250 , we will sequentially add their numbers to each other from right to left, in the order they go, and write down the result of their addition below. If the sum of the digits during addition exceeds 9 , then divide the sum by 10 , we write the remainder of the division under the current numbers, and move the integer part of the division to the left.

As a result, we get .

The most important rule with which we begin to study multiplication in a column:

Multiplication in a column by a two-digit number

Example: 46 times 73

This example can be written in a column.

Under the number 46 we write the number 73 according to the rule:

Units are written under units, and tens under tens

1 We start multiplying from units.

We multiply 3 by 6. It turns out 18.

• 18 units is 1 ten and 8 units.
• We write 8 units under the units, and remember 1 ten and add to the tens.

Now multiply 3 by 4 tens. Get 12.

12 tens, and even 1, only 13 tens.

There are no hundreds in this example, so we immediately write 1 in place of hundreds.

138 is first incomplete work.

2 We multiply tens.

Multiply 7 tens by 6 units to get 42 tens.

7 tens multiplied by 4 tens is 28 hundreds. 28 hundreds, and 4 more will make 32 hundreds.

There are no thousands in this example, so I immediately write 3 instead of thousands.

3220 is second incomplete work.

3 We add the first and second incomplete products according to the rule of addition in a column.

How to quickly multiply two-digit numbers in your head?

How to quickly multiply large numbers, how to master such useful skills? Most people have difficulty mentally multiplying two-digit numbers by single-digit numbers. And there is nothing to say about complex arithmetic calculations. But if desired, the abilities inherent in each person can be developed. Regular training, a little effort and the use of effective methods developed by scientists will achieve amazing results.

Choosing traditional methods

Proven decades methods of multiplying two-digit numbers do not lose their relevance. The simplest tricks help millions of ordinary schoolchildren, students of specialized universities and lyceums, as well as people involved in self-development, improve their computational skills.

Multiplication by factoring numbers

The easiest way to quickly learn to multiply large numbers in your head is to multiply tens and ones. First, tens of two numbers are multiplied, then ones and tens alternately. The four received numbers are summed up. To use this method, it is important to be able to memorize the results of multiplication and add them in your mind.

For example, to multiply 38 by 57, you need:

• split the number into (30+8)*(50+7) ;
• 30*50 = 1500 - memorize the result;
• 30*7 + 50*8 = 210 + 400 = 610 - remember;
• (1500 + 610) + 8*7 = 2110 + 56 = 2166

Naturally, it is necessary to know the multiplication table perfectly, since it will not be possible to quickly multiply in the mind in this way without the appropriate skills.

Multiplication in a column in the mind

The visual representation of the usual multiplication in a column is used by many in calculations. This method is suitable for those who can memorize auxiliary numbers for a long time and perform arithmetic operations with them. But the process is greatly simplified if you learn how to quickly multiply two-digit numbers by one-digit numbers. To multiply, for example, 47 * 81 you need:

• 47*1 = 47 - remember;
• 47*8 = 376 - we remember;
• 376*10 + 47 = 3807.

Remembering intermediate results will help pronouncing them out loud while summing up in your mind. Despite the complexity of mental calculations, after a short practice, this method will become your favorite.

The above methods of multiplication are universal. But knowing more efficient algorithms for some numbers will greatly reduce the number of calculations.

Multiply by 11

This is perhaps the easiest way and is used to multiply any two digit numbers by 11.

It is enough to insert their sum between the numbers of the multiplier:
13*11 = 1(1+3)3 = 143

If a number greater than 10 is obtained in brackets, then one is added to the first digit, and 10 is subtracted from the sum in brackets.
28*11 = 2 (2+8) 8 = 308

Multiplication of large numbers

It is very convenient to multiply numbers close to 100 by decomposing them into components. For example, you need to multiply 87 by 91.

• Each number must be represented as the difference between 100 and one more number:
  (100 - 13)*(100 - 9)
  The answer will consist of four digits, the first two of which are the difference between the first factor and the factor subtracted from the second bracket, or vice versa - the difference between the second factor and the factor subtracted from the first bracket.
  87 – 9 = 78
  91 – 13 = 78
• The second two digits of the answer are the result of multiplying those subtracted from two brackets. 13*9 = 144
• As a result, the numbers 78 and 144 are obtained. If, when writing the final result, a number of 5 digits is obtained, the second and third digits are summed up. Result: 87*91 = 7944 .

These are the easiest ways to multiply. After their repeated application, bringing the calculations to automatism, more complex techniques can be mastered. And after a while, the problem of how to quickly multiply two-digit numbers will cease to excite you, and memory and logic will improve significantly.

Mathematics lesson on the topic "Multiplication of three-digit numbers in a column." 3rd grade

A bad teacher teaches the truth, a good teacher teaches to find it.

The goal of modern Russian education has become the full-fledged formation and development of the student's abilities to independently outline an educational problem, formulate an algorithm for solving it, control the process and evaluate the result.
The new standard is distinguished by the implementation of a system-activity approach in teaching, where the position of the student is active, where he acts as an initiator and creator, and not a passive performer.

UUD formed in the lesson:

Personal:

• understanding the internal position of the student at the level of a positive attitude to the lesson
• moral and ethical assessment of digestible content
• adherence to moral standards and ethical requirements in behavior
• self-assessment based on the criterion of success

Communicative:

• planning learning collaboration with teacher and peers
• expressing one's thoughts with sufficient completeness and accuracy, using criteria to justify one's judgment

cognitive:

• extracting necessary information from tasks
• statement and formulation of the problem
• definition of primary and secondary information
• hypotheses and their justification

Regulatory:

• self-organization and organization of your workplace
• exercise of self-control
• fixing an individual difficulty in a trial educational action, the ability to predict

I. Organizing moment ( Presentation- slide 1)

Checking readiness for the lesson (slide 2)

- Check how your “workplace”, textbook, pencil case is organized.
Let's do finger exercises. (children touch their fingers with a neighbor on the desk and say):

Wish (thumb)
Large (medium)
Success (index)
All over (unnamed)
And everywhere (pinky)
Good luck! (whole palm)

Motivation for learning activities.

I also want to wish you good luck.
How do we start our work?

1. Encrypted word

- I offer you a very interesting task!
- What should be done?

Annex 1 (work in pairs)

- What was the word? (Success)
- Luck and success are waiting for each of you today at the lesson!
What is the largest three digit number. (124 ) (slide 3)
Tell me everything you know about this number. (It is natural, not round, it is in 124th place in the series of natural numbers, it is preceded by the number 123, followed by the number 125. The sum of the digits of this number is 7. It is three-digit. It has 1 hundred, 2 tens, 4 units)

2. Writing a number as a sum of bit terms

– Write it down as a sum of bit terms: 124 = 100 + 20 + 4 (slide 4)
- Swap notebooks with your desk mate and check each other's work.
- Now tell me, what do we know (can) about three-digit numbers?

II. Motivation

I know (I can) (slide 4)

• read
• write down
• compare
• represent as a sum of bit terms
• practice addition and subtraction
• practice multiplication and division

- What skills did we use when completing this task with the number 124? (Decompose three-digit numbers into the sum of bit terms)
Where can we use these skills? (When solving examples, for the convenience of calculations)
- Look at the blackboard.

800*3 200*4
412*2 123*3
112*4 300*3

What two groups can these expressions be divided into? (Expressions for multiplication of round and non-round three-digit numbers)
- An example of which column can we solve easily and quickly? Why? (First, we know how to multiply round numbers)
- Write in your notebook the answers of the examples of the first column.
- Who wrote it down, sit up straight. Check with a sample. (Slide 5)
Look at the examples of the second column. Can we solve these examples right away? Why? (No, we can't)

I want to know (slide 6)

– Would you like to know how to solve such examples? (How to perform multiplication of three-digit numbers in a column)
- What is the topic of today's lesson?

"Multiplication of three-digit numbers in a column" (slide 7)

What goals can we set? (Learn to multiply three-digit numbers in a column)
- Yes, that's right. With the multiplication of three-digit numbers in a column, you are not yet familiar!
- This is our main goal in the lesson!
- Make your guesses, how will we multiply a three-digit number by a one-digit number?

III. Finding a Solution

- What can help us not to make mistakes in solving examples? (Need ALGORITHM!)
- Now you need to work and correctly arrange the order of actions in the algorithm.
- We will be divided into two groups.
- The first group should restore the sequence of the algorithm, as you would have done in multiplication.
- With the second group, we will verbally analyze the algorithm of actions.
- The guys from the second group will evaluate the correctness of your algorithm. (Children line up in order)
- Read your algorithms, and now compare with the one I have on the slide. (slide 8)

ALGORITHM

1. WRITE.
2. I MULTIPLY THE UNITS.
3. WRITE UNITS UNDER UNITS.
4. I MULTIPLY TENS.
5. TENS WRITE UNDER TENS.
6. I MULTIPLY HUNDREDS.
7. WE WRITE HUNDREDS UNDER HUNDREDS.
8. READ THE ANSWER.

IV. Primary fastening

- And now we will use the algorithm and solve the examples of the second column (at the board with an explanation)

412 * 2 = 824
123 * 3 = 369
112 * 4 = 448

Did you like solving examples?
"Now let's get some rest."

IV. Fizminutka (slide 9)

- I will give tasks, and you will give the answer by the number of movements:

SO MANY TIMES STAMPING THE FOOT - 12: 3
SO MANY TIMES SHAP HANDS - 25: 5
WE WILL SIT SO MANY TIMES - 36: 9
WE LEAN NOW - 18: 3
WE JUMP EXACTLY SO MUCH - 36: 6
- RESTED? ON THE ROAD AGAIN.

V. Solution of the problem

– Can you use the skills acquired in the lesson when solving problems?
- Then let's decide!

(slide 10)

“The age of the birch under which the travelers built their hut is 121 years, and the age of the oak that grows nearby is 3 times more. How old is the oak tree? How many years is oak older than birch?
1) 121 * 3 \u003d 363 (g.) - the age of the oak.
2) 363 - 121 \u003d 242 (g.) - the difference.

Answer: The oak is 363 years old, the oak is 242 years older than the birch.

V. Independent work (slide 11)

– Can you solve the examples on your own?

223 * 3
212 * 4
241 * 2
313 * 3
413 * 2

- Swap notebooks and check if your neighbor solved the examples correctly.

VII. Reflection of educational activity in the lesson and the result of the lesson

What was our goal at the beginning of the lesson?
- Have you done it?

found out (an algorithm for multiplying three-digit numbers in a column) (slide 12)

- And where will this knowledge be useful to you? (At home, in the store.)
- Let's see how we worked, how you assessed your work and the work of the class.
- Now on the "ladder of mood" (slide 13) attach your star to the step that corresponds to your feelings, mood, state of your soul that you had throughout the lesson.

Multiplication of natural numbers by a column, examples, solutions.

Multiplication of natural numbers is conveniently carried out in a special way, which is called " multiplication by a column" or " column multiplication". The whole charm of this method lies in the fact that the multiplication of multi-valued natural numbers is reduced to the successive multiplication of two single-valued numbers.

In this article, we will analyze in the most detailed way the algorithm for multiplying two natural numbers by a column. We will describe the sequence of actions step by step, at the same time showing solutions to examples.

Page navigation.

What do you need to know to multiply natural numbers by a column?

Intermediate calculations when multiplying by a column are carried out using the multiplication table, so it is advisable to know it by heart so as not to waste time searching for the desired result.

Sooner or later, when multiplying by a column, we will encounter the multiplication of a single-digit natural number by zero. In this case, we will use the property of multiplying a natural number by zero: a 0=0, Where a is an arbitrary natural number.

We recommend that you deal with the material of the article column addition. This is due to the fact that at one of the stages of multiplication in a column, you have to add intermediate results (which are called incomplete products), using the principle of addition by a column.

Recording multipliers when multiplying in a column.

Let's start with the rules for writing multipliers when multiplying by a column.

The second multiplier is written under the first multiplier so that the first digits on the right that are different from the digit 0 are located one below the other. A horizontal line is drawn below the written multipliers, and a multiplication sign of the form “×” is placed on the left. Let us give examples of the correct notation of factors when multiplying by a column. The following shows the entries in the column of products of numbers 352 And 71 , 550 And 45 002 , and 534 000 And 4 300 .



Dealt with the record.

Now you can go directly to the process of multiplying two natural numbers by a column. First, consider the multiplication of a multi-digit number by a single-digit number. After that, we will analyze the multiplication by a column of two multivalued natural numbers.

Multiplication of a multi-valued natural number by a single-digit number by a column.

We will now bring column multiplication algorithm multi-valued natural number to a single-valued natural number. We will do this while describing the solution of the example.

Suppose we need to multiply a given multivalued natural number 45 027 for a given single number 3 .

We write the factors in the same way as multiplication by a column implies (in this case, the single-digit number is under the rightmost sign of the multi-digit number).

For our example, the entry will look like this:


Now we multiply the value of the units of a given multi-digit number by a given single-digit number. If we get a number less than 10 , then we write it under the horizontal line in the same column in which the given multiplied single-digit number is located. If we get a number 10 or a number greater than 10 , then under the horizontal line we write the value of the units digit of the resulting number, and remember the value of the tens digit (we add the memorized number to the multiplication result in the next step, after which we delete the memorized number from memory).

That is, we multiply 7 (this is the value of the units digit of the first multiplier 45 027 ) on 3 . We get 21 . Because 21 more 10 , then under the line we write the number 1 (this is the value of the units digit of the resulting number 21 ) and remember the number 2 (this is the value of the tens digit of the number 21 ). At this step, the entry will look like this:


Let's move on to the next step of the column multiplication algorithm. We multiply the value of the tens digit of a given multi-digit number by a given single-digit number and add to the product the number memorized at the previous stage (if we memorized it). If as a result we get a number less than ten, then we write it under the horizontal line to the left of the number already written there. If, as a result, we get the number ten or a number greater than ten, then under the horizontal line we write the value of the units digit of the resulting number, and remember the value of the tens digit (we will also use it in the next step).

So let's multiply 2 (this is the value of the tens digit of the first multiplier 45 027 ) on 3 , we have 6 . To this number we add the number memorized in the previous step 2 , we get 6+2=8 . Because 8 less than 10 , then under the horizontal line we write the number 8 to the desired position (at the same time, we do not need to remember any number, that is, now we have no numbers in memory). We have:


At the next step, we act similarly, but we are already multiplying the value of the hundreds digit of a given multi-digit number by a given single-digit natural number. We add to this work the memorized number (if it was memorized); compare result with number 10 ; if necessary, remember a new number, and write the desired number under the horizontal line to the left of the numbers already there.

Multiply 0 on 3 , we get 0 . Since we do not have any number in memory, then to the resulting number 0 nothing needs to be added. Number 0 less 10 , so we write 0 under the horizontal line at the desired position:


After that, we proceed to multiplying the value of the next digit of a given multi-valued natural number and a given single-valued natural number. We act in a similar way until we multiply the values ​​of all digits of a given multi-digit number by a given single-digit natural number.

So let's multiply 5 on 3 , we get 15 . Because 15>10 , then write under the line 5 and remember the number 1 :


Finally, we multiply 4 on 3 , we get 12 . TO 12 add the number memorized in the previous step 1 , we have 12+1=13 . Because 13 more than 10 , then write the number 3 to the right place and remember the number 1 :


Note that if at the last stage we had to remember a number, then it must be written under the horizontal line to the left of the numbers already there.

We have a number in memory 1 , so you need to write it in the right place under the line:


At this point, the process of multiplication by a column of a multi-valued natural number by a single-valued natural number ends, and the result of the multiplication is the number written under the horizontal line.

Thus, multiplication by a column of natural numbers 45 027 And 3 led us to the result 135 081 .

For clarity, we schematically depict the algorithm for multiplying a multi-valued natural number by a single-valued natural number by a column (this figure reflects only the general picture, but does not show all the nuances).



It remains to deal with multiplication by a column of a multi-valued natural number, in the record of which there is a digit on the right 0 or multiple numbers 0 consecutively, by a single number. We will also consider all the steps of multiplication by a column in such cases using an example. Moreover, we will take the numbers from the previous example, but in the entry of a multi-digit number we will add several digits 0 on right.

So let's multiply natural numbers 4 502 700 (we added two numbers 0 ) per number 3 .

In this case, we first write down the numbers to be multiplied in the way that multiplication by a column implies:


After that, we carry out multiplication by a column as if the numbers 0 right no.

Let's use the result from the example already solved above:


At the final stage of multiplication in a column under the horizontal line, to the right of the digits already there, we write down as many digits 0 , how many of them are on the right in the original multiplied number.

In our example, we need to add two digits 0 . The entry will look like:


This completes the column multiplication.

The result of multiplying a multivalued natural number 4 502 700 , whose record ends in zeros, to a single-digit natural number 3 is 13 508 100 .

Column multiplication of two multivalued natural numbers.

Let us describe all the stages of the algorithm for multiplying two multivalued natural numbers by a column.

The description will be carried out together with the solution of the example. Now we will assume that in the records of multiplied natural numbers there are no digits on the right 0 . The multiplication of multivalued natural numbers whose entries end in zeros will be considered at the end of this section.

Multiply by a column of numbers 207 on 8 063 .

We start by writing the multipliers under each other. Note that it is more convenient to place a multiplier on top, the record of which consists of a larger number of characters (in our example, we write the number on top 8 603 , since in his entry 4 sign, and the number 207 three-digit). If the multiplier records contain the same number of characters, then it does not matter which of the multipliers is written on top. So, we place the multipliers one below the other so that the numbers of the first multiplier are under the numbers of the second multiplier from right to left:


Now, at each next step, we will get the so-called incomplete works.

The first stage of the algorithm consists in multiplying the first factor by a column (in our example, this is the number 8 063 ) by the value of the units digit of the second multiplier (in our example, the value of the units digit of the number 207 is the number 7 ). All actions are similar to multiplying a multi-digit number by a single-digit number by a column (if necessary, return to the previous paragraph of this article), as a result, under the horizontal line we have the first incomplete product. At this stage, the entry will look like this:


We pass to the second stage. At this stage, we multiply the first factor by a column (in our example, this is the number 8 063 ) by the value of the tens place of the second factor, if it is not equal to zero. If the value of the tens digit of the second multiplier is equal to zero, then proceed to the next stage (in our example, the value of the tens digit of the number 207 is zero, so we will move on to the third step). We write the results under the line below the number already written there, starting from the position that corresponds to the tens place.

At the third, fourth and so on stages, we act similarly, multiplying the first factor by a column (number 8 063 ) by the value of the hundreds place of the second factor (if it is not equal to zero), then by the value of the thousand place (if it is not equal to zero), and so on. We write the results under the line below the numbers already written there, starting from the position corresponding to the digit of the single-digit number by which the multiplication is carried out at this stage.

So let's multiply the number 8 063 to the value of the hundreds place of the number 207 , that is, for the number 2 . We get the second incomplete product, and the solution of the example takes the following form:


So, all incomplete products are calculated. The last step of the algorithm remains, at which all incomplete products are added, and this is done in the same way as when adding in a column. Addition is performed using the existing record (incomplete products remain in the places where they are written, that is, they do not move anywhere), another horizontal line is drawn from below, a “+” sign is placed on the left, and the addition results are recorded under the bottom line . If there is only one number in the column, and there is no number stored in the memory at the previous stage, then it is written under the horizontal line.

In our example, we get:


The number formed below is the result of multiplying the original multi-valued natural numbers. So the product of numbers 8 063 And 207 equals 1 669 041 .

For clarity, we schematically depict the process of multiplication by a column of two natural numbers.



We will show the solution of another example for fixing the material.

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If we need to multiply natural numbers in the course of solving the problem, it is convenient to use a ready-made method for this, which is called "column multiplication" (or "column multiplication"). This is very convenient, since it can be used to reduce the multiplication of multi-digit numbers to the successive multiplication of single-valued ones.

Column Multiplication Basics

To conduct the calculation in a column, we will need a multiplication table. It is important to remember it by heart in order to count quickly and efficiently.

You will also need to remember what result we get when multiplying a natural number by zero. This is often seen in examples. We will need the property of multiplication, which is written in literal form as a 0 = 0 (a is any natural number).

To better understand how to multiply by a column, we recommend that you repeat the same addition method. One of the stages of calculations will be precisely the addition of intermediate results, and knowledge of this method will come in handy when adding numbers.

It is also important that you know how to compare natural numbers and remember what a place is.

As always, let's start with how to write the original numbers correctly. We need to take two factors and write them one below the other so that all non-zero numbers are located one below the other. Let's draw a horizontal line under them separating the answer, and add a multiplication sign on the left side.

Example 1

For example, to calculate and 71 , 550 45 002 and 534 000 4 300 , we write the following columns:

Next, we need to deal with the process of multiplication. First, let's see how to correctly multiply a multi-digit natural number by a single-digit one, and then we'll see how to multiply multi-digit numbers with each other.

If, in order to solve a problem, we need to multiply two natural numbers, one of which is single-valued and the second is multi-valued, then we can use the column method. To do this, we perform a sequence of steps, which we will explain immediately with an example. First, let's take a problem in which a multi-digit number has a digit other than zero at the end.

Example 2

Condition: calculate 45 027 3 .

Solution

Let's write the multipliers as the column multiplication method implies. We place the single-valued factor under the last sign of the multi-valued one. We received this entry:

Next, we need to perform sequential multiplication of the digits of a multi-digit number by the specified multiplier. If we get a number that is less than ten, we immediately enter it in the answer field under the horizontal line, strictly under the calculated digit. If the result was 10 or more, then we indicate under the required digit only the value of units from the resulting number, and remember the tens and add at the next step to the higher digit.

On specific numbers, the process will look like this:

1. We multiply 7 by 3 (we took the seven from the category of units of the first multi-valued factor): 7 3 \u003d 21. We got a number greater than ten, which means we write the number 1 from the right edge (the value of the unit digit of the number 21), and remember the two. Our entry becomes:

2. After that, we multiply the values ​​​​of the tens of the first factor by the second and add the two remaining from the previous stage to the result. If after that it turns out less than 10, then we enter the values ​​\u200b\u200bfor the corresponding digit, if more, we enter the value of one and transfer the tens further. In our example, we need to multiply 2 3 , it will be 6 . We add the tens remaining from the last multiplication (from the number 21, as we remember): 6 + 2 = 8. Eight is less than ten, which means that nothing needs to be transferred to the next digit. We write 8 in the right place and get:

3. Then we proceed in the same way. Now we need to multiply the values ​​of the hundreds place in the first multi-digit multiplier by the original single-digit one. The procedure is the same: if you memorized the number at the previous stage, add it to the result, compare it with ten and write it in the correct place.

Here you need to multiply 3 by 0 . According to the multiplication rules, the result will be 0 . We will not add anything, since at the previous stage the number was less than 10 . The resulting zero is also less than ten, so we write it in place under the horizontal line:

4. Go to the next category - multiply thousands. We continue the calculations according to the algorithm until the numbers in the multi-valued multiplier run out.

It remains to multiply 5 3 and get 15 . The result is greater than 10, write five and remember ten:

We only need to multiply 4 3 , it will be 12 . We add to the result the unit taken from the previous calculation. 13 is greater than 10 , we write 3 in the right place and save the unit.

We have no more digits left to multiply, but there is still one in stock. We will simply write it under the horizontal line to the left of all the numbers already there:

The process of counting with a column is now complete. We got a six-digit number, which is the correct solution to our problem.

Answer: 45,027 3 = 135,081.

To make it more clear, we presented the algorithm for multiplying a multivalued natural number by a single one in the form of a diagram. The essence of the counting process is correctly reflected here, but some nuances are not taken into account:

What if the condition of the problem contains a multi-digit number that ends with zero (or several zeros in a row)? Let's look at an example step by step. To make it easier, let's borrow the numbers from the previous problem and simply add a couple of zeros to the original multi-valued factor.

Solution

First, write the numbers in the right way.

After that, we carry out calculations, ignoring the zeros on the right. Let's take the results from the previous task so as not to count again:

The final step of the solution is to rewrite the zeros in the multi-digit number under the horizontal line in the result area. We need to add 2 extra zeros:

This number will be the answer to our problem. This completes the column multiplication.

Answer: 4 502 700 3 = 13 508 100 .

This method is quite suitable for those cases when both factors are multi-valued natural numbers. Let's analyze the process immediately with an example, as before. First, let's take numbers without zeros at the end, and then consider entries with zeros.

Example 4

Condition: calculate how much will be 207 8 063 .

Solution

Let's start, as always, with the correct notation of factors. More convenient is the way of writing, in which the multiplier with a large number of signs is on top. So let's write 8063 first and then 207 below it. If the number of characters in the factors is the same, then the order of recording does not matter. In our problem, we need to place the numbers of the first factor under the numbers of the second from right to left:

We begin to sequentially multiply the values ​​​​of the digits. In this case, we will get results that are called incomplete products.

1. The first step is that we need to multiply the values ​​of the units in the first and second multiplier. In our case, these are 3 and 7 . We do everything in the same way as we already explained in the previous paragraph (if necessary, read it again). As a result, we get the first incomplete product, which is an intermediate result:

2. The second step is to multiply the tens values. We multiply the first multiplier by a column by the value of the tens digit of the second multiplier (provided that it is not equal to 0). We write the result under the line under the tens place. If in the second multiplier there is 0 in place of tens, then we immediately proceed to the next stage.

3. Follow the next steps in the same way, multiplying in turn the values ​​of the required digits (if they are not equal to 0). We enter the results below the line.

So, we need to multiply 8,063 by the hundreds of values ​​in 207 (i.e. two). We have received the second incomplete product, we write it like this:

We got all the incomplete works we needed. Their number is equal to the number of digits in the second multiplier (except 0). The last thing left for us to do is add the two works in a column using the same notation. We do not rewrite the numbers anywhere: they remain with the same shift to the left. We underline them with an additional horizontal line and put a plus on the left. We add according to the already studied rules for addition in a column (remember the tens if the number turned out to be more than 10, and add them in the next step). Our task will be:

The seven-digit number obtained under the line is the result of multiplying the original natural numbers we need.

Answer: 8063 207 = 1669041.

The process of multiplying two multi-valued numbers of columns can also be represented as a visual diagram:

To better consolidate the material, we give the solution of another example.

Example 5

Condition: multiply 297 by 321.

Solution

We start with the correct notation of the multipliers. The number of characters in them is the same, so the order of writing does not really matter:

1. The first stage - we multiply 297 by 1, which is in the category of units of the second multiplier.

2. Then we multiply in the same way the first factor by 2, which is in tens of the second factor. We get the second incomplete product.