Rational ways of making decisions

The current level of development of computing automation tools has created an illusion among many that developing computing skills is not at all necessary. This affected the preparedness of schoolchildren. In the absence of a calculator, even simple computational tasks become a problem for many.

At the same time, exam tasks and materials for the Unified State Exam contain many tasks, the solution of which requires from the subjects the skills of rational organization of calculations.

In this article, we will consider some ways to optimize calculations and their application for competitive problems.

Most often, methods of optimizing calculations are associated with the application of the basic laws of performing arithmetic operations.

For example:

125 24 \u003d 125 8 3 \u003d 1000 3 \u003d 3000; or

98 16 (100 - 2) 16 \u003d 100 16 - 2 16 \u003d 1600 - 32 \u003d 1568, etc.

Another direction - use of abbreviated multiplication formulas.

96 104 \u003d (100 - 4) (100 + 4) \u003d 100 2 - 4 2 \u003d 10000 - 16 \u003d 9984; or

115 2 \u003d (100 + 15) 2 \u003d 10000 + 2 15 100 + 225 \u003d 10525.

The following example is interesting for calculations.

Calculate:

(197 · 203 + 298 · 302 + 13) / (1999 · 2001 + 2993 · 3007 + 50) =
= ((200 – 3) · (200 + 3) + (300 – 2) · (300 + 2) + 13) / ((2000 – 1) · (2000 + 1) + (3000 – 7) · (3000 + 7) + 50) =
= (200 2 – 3 2 + 300 2 – 2 2 + 13) / (2000 2 – 1 2 + 3000 2 – 7 2 – 50) =
= 130000 / 13000000 = 0,01

These are almost standard ways to optimize computations. Sometimes more exotic ones are offered. As an example, consider a method for multiplying two-digit numbers, the sum of their units is 10.

54 26 \u003d 50 30 + 4 (26 - 50) \u003d 1500 - 96 \u003d 1404 or

43 87 \u003d 40 90 + 3 (87 - 40) \u003d 3600 + 141 \u003d 3741.

The multiplication scheme can be understood from the figure.

Where does this multiplication scheme come from?

Our numbers by condition have the form: M \u003d 10m + n, K \u003d 10k + (10 - n). Let's compose a work:

M K \u003d (10m + n) (10k + (10 - n)) \u003d
\u003d 100mk + 100m - 10mn + 10nk + 10n - n 2 \u003d
\u003d m (k + 1) 100 + n (10k + 10 - n) \u003d
\u003d (10m) (10 (k + 1)) + n (K - 10m) and the method is justified.

There are many clever ways to turn fairly complex calculations into oral problems. But you can't think that everyone needs to memorize these and a bunch of other ingenious ways to simplify calculations. It is only important to learn some of the basic ones. The analysis of others makes sense only to develop skills in applying basic methods. It is their creative application that makes it possible to quickly and correctly solve computational problems.

Sometimes, when solving calculation examples, it is convenient to go from transforming an expression with numbers to transforming polynomials. Consider the following example.

Calculate in the most rational way:

3 1/117 4 1/110 -1 110/117 5 118/119 - 5/119

Decision.

Let a \u003d 1/117 and b \u003d 1/119. Then 3 1/117 \u003d 3 + a, 4 1/119 \u003d 4 + b, 1 116/117 \u003d 2 - a, 5 118/119 \u003d 6 - b.

Thus, the given expression can be written as (3 + a) (4 + b) - (2 - a) (6 - b) - 5b.

After performing simple transformations of the polynomial, we get 10a or 10/117.

Here we get that the value of our expression does not depend on b. And this means that we calculated not only the value of this expression, but also any other one obtained from (3 + a) (4 + b) - (2 - a) (6 - b) - 5b by substituting the values \u200b\u200bof a and b. If, for example, a \u003d 5/329, then in the answer we get 50 / 329 whatever b.

Consider another example, the solution of which with the help of a calculator is almost impossible, and the answer is quite simple if you know the approach to solving examples of this type

Calculate

1/6 7 1024 - (7 512 + 1) (7 256 + 1) (7 128 + 1) (7 64 + 1) (7 32 + 1) (7 16 + 1) ( 7 8 + 1) (7 4 + 1) (7 2 + 1) (7 + 1)

Decision.

We transform the condition

1/6 7 1024 - 1/6 (7 512 + 1) (7 256 + 1) (7 128 + 1) (7 64 +1) (7 32 + 1) (7 16 + 1) (7 8 + 1) (7 4 + 1) (7 2 + 1) (7 + 1) (7 - 1) \u003d

1/6 7 1024 - 1/6 (7 512 + 1) (7 256 + 1) (7 128 + 1) (7 64 + 1) (7 32 + 1) (7 16 + 1 ) (7 8 + 1) (7 4 + 1) (7 2 + 1) (7 2 - 1) \u003d

1/6 7 1024 - 1/6 (7 512 + 1) (7 256 + 1) (7 128 + 1) (7 64 + 1) (7 32 + 1) (7 16 + 1 ) (7 8 + 1) (7 4 + 1) (7 4 - 1) \u003d

1/6 7 1024 - 1/6 (7 512 + 1) (7 256 + 1) (7 128 + 1) (7 64 + 1) (7 32 + 1) (7 16 + 1 ) (7 8 + 1) (7 8 - 1) \u003d

1/6 7 1024 - 1/6 (7 512 + 1) (7 256 +1) (7 128 + 1) (7 64 + 1) (7 32 + 1) (7 16 + 1 ) · (7 16 - 1) \u003d… \u003d

1/6 7 1024 - 1/6 (7 512 + 1) (7 512 - 1) \u003d 1/6 7 1024 - 1/6 (7 1024 - 1) \u003d 1/6

Consider one of the examples that has already become textbook in the examination materials for the course of the basic school.

Calculate the amount:

1/2 + 1 / (2 · 3) + 1 / (3 · 4) + 1 / (4 · 5) + ... + 1 / (120 · 121) \u003d

= (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1/4) + (1/4 – 1/5) + … + (1/120 – 1/121) =

= 1 – 1/121 = 120/121.

That is, the solution to this problem was made possible by the method of replacing each fraction with the difference of two fractions. The sum turned out to be a pair of opposite numbers to all except the first and the last.

But this example can be generalized. Consider the amount:

k / (n (n + k)) + k / ((n + k) (n + 2k)) + k / ((n + 2k) (n + 3k)) +… + k / (( n + (m – 1) k) (n + mk))

For it, all the same arguments are valid as in the previous example. Indeed:

1 / n – 1 / (n + k) \u003d k / (n (n + k));

1 / ((n + k) – 1 / (n + 2k) \u003d k / ((n + k) (n + 2k)), etc.

Then we construct the answer according to the same scheme: 1 / n – 1 / (n + mk) \u003d mk / (n (n + mk))

And more about the "long" amounts.

Amount

X \u003d 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 + 1/1024

can be calculated as the sum of 11 terms of a geometric progression with the denominator 1/2 and the first term 1. But the same sum can be calculated by a 5th grade student who has no idea about progressions. To do this, it is enough to successfully choose a number that we add to the sum of X. This number here will be 1/1024.

We calculate

X + 1/1024 \u003d 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 + (1/1024 + 1 / 1024) \u003d
= 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 + 1/512 =
=1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/256 = … = 1 + 1/2 + 1/2 = 2.

It is now obvious that X \u003d 2 – 1/1024 = 1 1023 / 1024 .

The second method is no less promising. It can be used to calculate the amount:

S \u003d 9 + 99 + 999 + 9999 +… + 99 999 999 999.

Here the "lucky" number is 11. Adding it to S and distributing it evenly among all 11 terms. Each of them will then get 1. Then we have:

S + 11 \u003d 9 + 1 + 99 + 1 + 999 + 1 + 9999 + 1 +… + 99 999 999 999 + 1 \u003d
= 10 + 100 + 1000 + 10000 + ... + 100 000 000 000 = 111 111 111 110;

Therefore, S \u003d 111 111 111 110 – 11 = 111 111 111 099.

1 + 11 + 111 + 1111 + ... + 1 111 111 111?

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In the distant past, when the number system was not yet invented, people counted everything on their fingers. With the advent of arithmetic and the foundations of mathematics, it has become much easier and more practical to keep track of goods, products, and household items. However, what does the modern system of calculus look like: what types are existing numbers divided into and what does "rational form of numbers" mean? Let's figure it out.

How many varieties of numbers are there in mathematics?

The very concept "number" designates a certain unit of any object that characterizes its quantitative, comparative or ordinal indicators. In order to correctly count the number of certain things or to carry out some mathematical operations with numbers (add, multiply, etc.), first of all, you should familiarize yourself with the varieties of these same numbers.

So, the existing numbers can be divided into the following categories:

1. Natural numbers are those numbers with which we count the number of objects (the smallest natural number is 1, it is logical that the series of natural numbers is infinite, that is, there is no greatest natural number). The set of natural numbers is usually denoted by the letter N.
2. Whole numbers. This set includes all while adding negative values \u200b\u200bto it, including the number "zero". The designation of the set of integers is written in the form of the Latin letter Z.
3. Rational numbers are those that we can mentally transform into a fraction, the numerator of which will belong to the set of integers, and the denominator - natural numbers. Below we will take a closer look at what "rational number" means, and give some examples.
4. - a set that includes all rational and This set is denoted by the letter R.
5. Complex numbers contain part of the real and part of the variable number. They are used in solving various cubic equations, which, in turn, can have negative expressions in formulas (i 2 \u003d -1).

What does "rational" mean: we look at examples

If the numbers that we can represent as an ordinary fraction are considered rational, then it turns out that all positive and negative integers are also included in the set of rational ones. After all, any integer, for example 3 or 15, can be represented as a fraction, where the denominator will be one.

Fractions: -9/3; 7/5, 6/55 are examples of rational numbers.

What does "rational expression" mean?

Move on. We have already figured out what the rational form of numbers means. Let's now imagine a mathematical expression that consists of the sum, difference, product, or quotient of various numbers and variables. Here is an example: a fraction, in the numerator of which the sum of two or more integers, and the denominator contains both an integer and a certain variable. It is this expression that is called rational. Based on the rule "you cannot divide by zero", one can guess that the value of this variable cannot be such that the value of the denominator turns to zero. Therefore, when solving a rational expression, you must first determine the range of the variable. For example, if the denominator is x + 5-2, then it turns out that "x" cannot be -3. Indeed, in this case, the entire expression turns to zero, therefore, when solving it, it is necessary to exclude the integer -3 for this variable.

How to solve rational equations correctly?

Rational expressions can contain a fairly large number of numbers and even 2 variables, so sometimes their solution becomes difficult. To facilitate the solution of such an expression, it is recommended to perform certain operations in a rational way. So what does it mean "in a rational way" and what rules should be applied in the decision?

1. The first kind, when it is enough just to simplify the expression. To do this, you can resort to the operation of reducing the numerator and denominator to an irreducible value. For example, if the numerator contains the expression 18x, and the denominator is 9x, then, reducing both indicators by 9x, we get just an integer equal to 2.
2. The second method is practical when we have a monomial in the numerator and a polynomial in the denominator. Let's take an example: in the numerator we have 5x, and in the denominator - 5x + 20x 2. In this case, it is best to put the variable in the denominator outside the brackets, we get the following form of the denominator: 5x (1 + 4x). Now you can use the first rule and simplify the expression by cutting 5x in the numerator and denominator. As a result, we get a fraction of the form 1/1 + 4x.

What can you do with rational numbers?

The set of rational numbers has a number of peculiarities. Many of them are very similar to the characteristic present in integers and natural numbers, since the latter are always included in the set of rational numbers. Here are some properties of rational numbers, knowing which, you can easily solve any rational expression.

1. The commutative property allows you to sum two or more numbers, regardless of their order. Simply put, the sum does not change from a change in the places of the terms.
2. The distributive property allows solving problems using the distribution law.
3. And finally, the operations of addition and subtraction.

Even schoolchildren know what "rational form of numbers" means and how to solve problems based on such expressions, so an educated adult simply needs to remember at least the basics of the set of rational numbers.

A rational way of making decisions in general form can be represented as follows.

The use of an administrative method of decision-making is expressed in the fact that the manager explores alternatives until he finds a satisfactory solution, that is, ensuring the achievement of the goal at a minimum level. He chooses the first alternative that meets his goals. This choice is limited by the values, experience and level of training of the leader. If the manager does not have alternatives that meet the minimum level of the set goals, he reduces the value of this level and accepts the first alternative. He is guided only by the specific circumstances of the situation and his powers.

With an intuitive way of making a decision, there is no systematic approach to the choice of alternatives. This method is often used by creative people. Research shows that the characteristics of these individuals include a great need for independence, business egoism, erudition, broad interests. This does not mean that only such leaders are creative individuals. They can also be those who use other ways of making decisions. The intuitive form occurs when a decision is made by chance. Most decisions are justified using a combination of rational and intuitive methods.

Who should make the decision: an individual or a group? There are several possible schemes: 1) the leader can make a decision alone; 2) the decision can be made by the leader after consultation with others; 3) those who are influenced by the decision can accept it as a group (the leader in this case acts as one of the members of the group). In all cases, it is important to follow the established procedures, the implementation of which ensures the necessary validity and reliability of a particular decision (Table 16.4).

Group decision-making ensures the participation of those concerned and increases their willingness to consciously implement the decision. Coordination of subsequent work is facilitated, communications are improved, the variety of alternatives considered increases, and the amount of information used is expanded. At the same time, the literature on management also notes possible disadvantages of group decision-making: it can be longer, groups may be less decisive and more often compromise, often fall under someone's influence, individuals can use the group to increase their influence;

sometimes groups cannot make a decision at all due to internal conflicts and disagreements.

Groups are best used for decision making when accuracy is critical. Efficiency is more important in some situations, accuracy in others. The group is often more accurate than the individual. Equally important is group cohesion, with a recognized coordinating leadership role. There are many situations where a solution requires many skills and experience that cannot be inherent in one person

On the basis of scientific research and extensive practice of making managerial decisions in recent decades, a number of methods of group decision-making have been developed, which have sharply increased the objectivity and validity of this process. Among them are brainstorming, the nominal group method, the Delphi method.

Brainstorming is undertaken by a group as an idea-generating process when all possible alternatives are considered critically.

The nominal group method limits discussion or communication with each other to a certain limit. Members of the group will attend the meeting and will act independently. The problem is posed first, and then the next steps are taken.

1. Before the discussion begins, everyone independently writes down their ideas for the submitted problem.

2. All ideas are recorded by each member of the group.

3. The group discusses ideas in order to clarify and evaluate them.

4. Each member of the group independently determines the importance of all ideas. The final decision is defined as the idea with the highest cumulative rating.

The main advantage of this method is that it allows the group to formally hold a common meeting, but does not limit the independence of thinking of each.

The most difficult and time consuming is the use of the Delphi method. It is similar to the nominal group method, with the difference that the physical presence of all group members is not required. The Delphi Method does not involve group members having to meet each other face to face. This method is characterized by the following steps.

1. The problem is determined; group members are asked to provide possible solutions by responding to a carefully designed questionnaire.

2. Each member of the group answers the first questionnaire anonymously and independently.

3. The results of the first questionnaire are collected in the center, deciphered and summarized.

4. Each team member receives a copy of the results.

5. After reviewing the results, the experts are asked to give their solutions again. As a rule, new solutions are given or changes in the initial position appear.

6. These steps are repeated as often as necessary until a consensus is reached.

The advantage of the method is the independence of the opinion of experts located at a spatial distance from each other.

In between group and individual decision-making is the way in which the manager constantly resorts to the help of qualified consultants before making a decision. He understands the need for consultation and knows how to use the potential of the group for an informed and timely solution of an urgent issue.

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Kozhinova Anastasia

MUNICIPAL NON-TYPE BUDGETARY

EDUCATIONAL INSTITUTION

"LYCEUM №76"

WHAT IS THE SECRET OF A RATIONAL ACCOUNT?

Performed:

Pupil 5 "B" class

Kozhinova Anastasia

Leader:

Mathematic teacher

Shchiklina Tatiana

Nikolaevna

Novokuznetsk 2013

Introduction ………………………………………………………… 3

Main part .... ……………………………………… .......... 5-13

Conclusion and conclusions ……………………………… ............... 13-14

References ……………………………………… .................. 15

Appendices ……………………………………………………. 16-31

I... Introduction

Problem: find the values \u200b\u200bof numeric expressions

Objective: search, study of existing methods and techniques of rational calculation, their application in practice.

Tasks:

1. Conduct a mini research in the form of a questionnaire among parallel classes.

2. Analyze on the topic of research: literature available in the school library, information in the academic textbook on mathematics for grade 5, on the Internet.

3. Choose the most effective methods and means of rational counting.

4. Conduct a classification of existing techniques for rapid oral and written counting.

5. Create memos containing the techniques of rational counting for use in parallel 5 classes.

Object of study: rational account.

Subject of study: ways of rational counting.

For the effectiveness of my research work, I used the following methods: analysis of information obtained from various resources, synthesis, generalization; poll in the form of a survey. The questionnaire was developed by me in accordance with the purpose and objectives of the study, the age of the respondents and is presented in the main part of the work.

In the course of the research work, issues related to the methods and techniques of rational calculation were considered, and recommendations were given to eliminate problems with computing skills, to form a computing culture.

II... Main part

Formation of the computational culture of students

5-6 grades.

Obviously, the methods of rational counting are a necessary element of the computational culture in the life of every person, first of all, the strength of their practical significance, and students need it in almost every lesson.

Computational culture is the foundation of the study of mathematics and other academic disciplines, because, in addition to the fact that calculations activate memory, attention, help to rationally organize activities and significantly affect human development.

In everyday life, in classrooms, when every minute is valued, it is very important to quickly and efficiently carry out oral and written calculations, without making mistakes and without using any additional computing means.

We, schoolchildren, are faced with this problem everywhere: in class, at home, in a store, etc. In addition, after grades 9 and 11, we will have to take exams in the form of IGA and USE, where the use of a microcalculator is not allowed. Therefore, the problem of forming a computational culture in every person becomes extremely important, an element of which is mastering the methods of rational calculation.

It is especially necessary to master the techniques of rational counting.

in the study of subjects such as mathematics, history, technology, computer science, etc., that is, rational counting helps to master related subjects, better navigate the material being studied, in life situations. So what are we waiting for? Let's go into the world of secrets of Rational counting techniques !!!

What problems do learners have when performing calculations?

Often, peers of my age have problems when performing various tasks in which they need to make calculations quickly and in a convenient way ... Why???

Here are some suggestions:

1. The student has a poor understanding of the topic studied topic

2. The student does not repeat the material

3. The student has poor numeracy skills

4. The student does not want to study the topic

5. The student believes that it will not be useful to him.

All these assumptions I took from my experience and the experience of my classmates and peers. However, in exercises of a computational nature, the skills of rational counting play an important role, so I have studied, apply and want to present you with some techniques of rational counting.

Rational methods of oral and written calculations.

In work and everyday life, the need for various kinds of calculations constantly arises. Using the simplest methods of verbal counting reduces fatigue, develops attention and memory. The use of rational calculation methods is necessary to increase labor, accuracy and speed of calculations. The speed and accuracy of calculations can be achieved only with the rational use of methods and means of mechanization of calculations, as well as with the correct use of methods of oral counting.

I... Simplified addition of numbers

There are four known methods of addition to speed up the calculations.

Bitwise sequential addition method is used in oral calculations, as it simplifies and speeds up the summation of terms. When using this method, addition begins with the highest digits: the corresponding digits of the second term are added to the first term.

Example. Find the sum of the numbers 5287 and 3564 using sequential bitwise addition.

Decision. We will make the calculation in the following sequence:

5 287 + 3 000 = 8 287;

8 287 + 500 = 8 787;

8 787 + 60 = 8 847;

8 847 + 4 = 8 851.

Answer: 8 851. (combination-transposition law)

Another way of sequential bitwise addition consists in the fact that the higher category of the second term is added to the highest category of the first term, then the next category of the second term is added to the next category of the first term, etc.

Consider this solution using the given example, we get:

5 000 + 3 000 = 8 000;

200 + 500 = 700;

Answer: 8851.

Round number method ... A number that has one significant digit and ends with one or more zeros is called a round number. This method is used when one can choose from two or more terms that can be completed to a round number. The difference between the round number and the number specified in the calculation condition is called the complement. For example, 1,000 - 978 \u003d 22. In this case, 22 is the complement of 978 to 1,000.

To make addition in the way of a round number, it is necessary to round one or more terms close to round numbers, add round numbers and subtract arithmetic additions from the resulting sum.

Example. Find the sum of the numbers 1 238 and 193 using the round number method.

Decision. Let's round the number 193 to 200 and add as follows: 1 238 + 193 \u003d (1 238 + 200) - 7 \u003d 1 431. (combination law)

Grouping of terms ... This method is used in the case when the terms, when grouped together, add up to round numbers, which are then added together.

Example. Find the sum of the numbers 74, 32, 67, 48, 33 and 26.

Decision. Let's sum up the numbers grouped as follows: (74 + 26) + (32 + 48) + (67 + 33) \u003d 280.

(combination-displacement law)

or, when the numbers are grouped together, the same amounts are obtained:

Example: 1 + 2 + 3 + 4 + 5 +… + 97 + 98 + 99 + 100 \u003d (1 + 100) + (2 + 99) + (3 + 98) +… \u003d 101x50 \u003d 5050

(combination-displacement law)

II... Simplified number subtraction techniques

A method of sequential bitwise subtraction. This method sequentially subtracts each digit subtracted from the decremented one. It is used when numbers cannot be rounded.

Example. Find the difference between the numbers 721 and 398.

Decision. Let's perform the steps to find the difference of the given numbers in the following sequence:

we represent the number 398 as a sum: 300 + 90 + 8 \u003d 398;

let's perform a bitwise subtraction:

721 - 300 = 421; 421 - 90 = 331; 331 - 8 = 323.

Round number method ... This method is used when the subtracted is close to a round number. For the calculation, it is necessary to subtract the subtracted, taken as a round number, from the reduced, and add the arithmetic addition to the resulting difference.

Example... Let's calculate the difference of numbers 235 and 197 using the round number method.

Decision. 235 - 197 \u003d 235 - 200 + 3 \u003d 38.

III... Simple multiplication techniques

Multiplication by one followed by zeros. When a number is multiplied by a number that includes one followed by zeros (10; 100; 1,000, etc.), as many zeros are assigned to it on the right as there are in the factor after one.

Example. Find the product of the numbers 568 and 100.

Decision. 568 x 100 \u003d 56 800.

Serial bitwise multiplication method ... This method is used when multiplying a number by any single digit. If you need to multiply a two-digit (three-, four-digit, etc.) number by a one-digit number, then first the one-digit factor is multiplied by tens of another factor, then by its units and the resulting products are summed up.

Example. Find the product of the numbers 39 and 7.

Decision. 39 x 7 \u003d (30 + 9) x 7 \u003d (30 x 7) + (9 x 7) \u003d 210 + 63 \u003d 273. (distribution law of multiplication with respect to addition)

Round number method ... Use this method only when one of the factors is close to a round number. The multiplier is multiplied by a round number, and then by the arithmetic's complement and at the end the second is subtracted from the first product.

Example. Find the product of the numbers 174 and 69.

174 x 69 \u003d 174 x (70-1) \u003d 174 x 70 - 174 x 1 \u003d 12 180 - 174 \u003d 12 006. (the distribution law of multiplication relative to subtraction)

A way to decompose one of the factors. In this method, one of the factors is first decomposed into parts (terms), then the second factor is alternately multiplied by each part of the first factor and the resulting products are summed up.

Example... Find the product of the numbers 13 and 325.

Let us expand the number 13 into terms: 13 \u003d 10 + 3. Multiply each of the resulting terms by 325: 10 x 325 \u003d 3 250; 3 x 325 \u003d 975. Summing up the resulting products: 3 250 + 975 \u003d 4 225

Learning the skills of rational oral counting will make your work more efficient. This is possible only with a good mastery of all the above arithmetic operations. The use of rational counting techniques speeds up calculations and provides the required accuracy. But not only you need to be able to calculate, but you also need to know the multiplication table, the laws of arithmetic operations, classes and categories.

Oral counting systems exist that allow you to count orally quickly and efficiently. We will look at some of the most commonly used techniques.

1. Multiply a two-digit number by 11.

We studied this method, but we did not fully study it. the secret of this method is that it can be calculated by the laws of arithmetic operations.

Examples:

23x11 \u003d 23x (10 + 1) \u003d 23x10 + 23x1 \u003d 253 (distribution law of multiplication relative to addition)

23x11 \u003d (20 + 3) x 11 \u003d 20x11 + 3x11 \u003d 253 (distribution law and round number method)

We studied this method, but we did not know another the secret of multiplying two-digit numbers by 11.

Observing the results obtained when multiplying two-digit numbers by 11, I noticed that you can get the answer in a more convenient way : when multiplying a two-digit number by 11, the digits of this number are moved apart and the sum of these digits is put in the middle.

a) 23 11 \u003d 253, since 2 + 3 \u003d 5;

b) 45 11 \u003d 495, since 4 + 5 \u003d 9;

c) 57 11 \u003d 627, because 5 + 7 \u003d 12, two were placed in the middle, and one was added to the category of hundreds;

d) 78 11 \u003d 858, since 7 + 8 \u003d 15, then the number of tens will be 5, and the number of hundreds will increase by one and will be equal to 8.

I found confirmation of this method on the Internet.

2) The product of two-digit numbers, which have the same number of tens, and the sum of the ones is 10, that is, 23 27; 34 36; 52 58 etc.

The rule: the figure of tens is multiplied by the next figure in the natural row, the result is written down and the product of units is attributed to it.

a) 23 27 \u003d 621. How did you get 621? We multiply the number 2 by 3 (the “two” is followed by the “three”), it will be 6, and next to it we will add the product of ones: 3 7 \u003d 21, it turns out 621.

b) 34 36 \u003d 1224, since 3 4 \u003d 12, we assign 24 to the number 12, this is the product of units of these numbers: 4 6.

c) 52 58 \u003d 3016, since we multiply the tens digit 5 \u200b\u200bby 6, it will be 30, we attribute the product of 2 and 8, that is, 16.

d) 61 69 \u003d 4209. It is clear that 6 multiplied by 7 and got 42. And where does zero come from? The units were multiplied and got: 1 9 \u003d 9, but the result must be two-digit, so we take 09.

3) Division of three-digit numbers, consisting of the same digits, by the number 37. The result is the sum of these equal digits of a three-digit number (or a number equal to three times the digit of a three-digit number).

Examples: a) 222: 37 \u003d 6. This is the sum of 2 + 2 + 2 \u003d 6; b) 333: 37 \u003d 9, since 3 + 3 + 3 \u003d 9.

c) 777: 37 \u003d 21, i.e. 7 + 7 + 7 \u003d 21.

d) 888: 37 \u003d 24, since 8 + 8 + 8 \u003d 24.

We also take into account that 888: 24 \u003d 37.

III... Conclusion

To solve the main secret in the topic of my work, I had to work hard - to search, analyze information, question classmates, repeat the early known methods and find many unfamiliar ways of rational counting, and, finally, to understand what is his secret? And I realized that the main thing is to know and be able to apply the known, find new rational methods of counting, the multiplication table, the composition of the number (classes and categories), the laws of arithmetic operations. Besides,

look for new ways to do this:

- Simplified addition of numbers: (method of sequential bitwise addition; method of a round number; method of decomposing one of the factors into terms);

-Simplified number subtraction techniques (sequential bitwise subtraction method; round number method);

-Simple multiplication techniques (multiplication by one followed by zeros; method of sequential bitwise multiplication; method of a round number; method of decomposition of one of the factors ;

- Secrets of quick verbal counting (multiplying a two-digit number by 11: when multiplying a two-digit number by 11, the digits of this number are pushed apart and put in the middle the sum of these digits; the product of two-digit numbers, which have the same number of tens, and the sum of ones is 10; Division of three-digit numbers consisting of the same digits, to the number 37. There are probably many more such ways, so I will continue to work on this topic next year.

IV. List of references

1. Savin A.P. Mathematical miniatures / A.P. Savin. - M .: Children's literature, 1991

2. Zubareva II, Mathematics, grade 5: a textbook for students of educational institutions / II Zubareva, A.G. Mordkovich. - M .: Mnemosina, 2011

4.http: / / www. xreferat.ru

5.http: / / www. biografia.ru

6.http: / / www. Mathematics-repetition. ru

V... Applications

Mini research (survey in the form of a questionnaire)

To identify students' knowledge about rational counting, I conducted a survey in the form of a questionnaire on the following questions:

* Do you know what rational counting techniques are?

* If yes, then where, and if not, why?

* How many ways of rational counting do you know?

* Do you have difficulty in verbal counting?

* How do you study in math? a) by "5"; b) by "4"; c) to "3"

* What do you like best about math?

a) examples; b) tasks; c) fractions

* Where do you think oral counting can come in handy, besides mathematics? * Do you remember the laws of arithmetic operations, if so which ones?

After conducting a survey, I realized that my classmates do not know enough the laws of arithmetic operations, most of them have problems with rational counting, many students count slowly and with errors, and everyone wants to learn how to count quickly, correctly and in a convenient way. Therefore, the topic of my research work is extremely important for all students and not only.

1. Interesting oral and written methods of calculations that we studied in mathematics lessons, using the examples of the textbook "mathematics, grade 5":

Here are some of them:

to quickly multiply a number by 5, it is enough to note that 5 \u003d 10: 2.

For example, 43x5 \u003d (43x10): 2 \u003d 430: 2 \u003d 215;

48x5 \u003d (48: 2) x10 \u003d 24x10 \u003d 240.

To multiply the number by 50 , you can multiply it by 100 and divide by 2.

For example: 122x50 \u003d (122x100): 2 \u003d 12200: 2 \u003d 6100

To multiply the number by 25 , you can multiply it by 100 and divide by 4,

For example, 32x25 \u003d (32x100): 4 \u003d 3200: 4 \u003d 800

To multiply the number by 125 , you can multiply it by 1000 and divide by 8,

For example: 192x125 \u003d (192x1000): 8 \u003d 192000: 8 \u003d 24000

To divide a round number with two 0s by 25 , you can divide it by 100 and multiply it by 4.

For example: 2400: 25 \u003d (2400: 100) x 4 \u003d 24 x 4 \u003d 96

To divide a round number by 50 , can be divided by 100 and multiplied by 2

For example: 4500: 50 \u003d (4500: 100) x 2 \u003d 45 x 2 \u003d 90

But not only you need to be able to calculate, but you also need to know the multiplication table, the laws of arithmetic operations, the composition of the number (classes and categories) and have the skills to use them

The laws of arithmetic operations.

a + b = b + a

The displacement law of addition

(a + b) + c = a + (b + c)

Combination law of addition

a · b = b · a

The travel law of multiplication

(a · b) · c = a · (b · c)

Combination law of multiplication

(a = b) · c = a · c = b · c

Distribution law of multiplication (relative to addition)

Multiplication table.

What is multiplication?

It's a smart addition.

After all, it is smarter to multiply once,

Than putting it all together for an hour.

Multiplication table

All of us will be useful in life.

And not without reason is it named

By MULTIPLICATION she is!

Ranks and classes

In order to make it convenient to read, as well as to memorize numbers with large values, they should be divided into so-called "classes": starting from the right, the number is separated by a space into three digits "first class", then three more digits, "second class" and etc. Depending on the meaning of the number, the last class can end with either three or two or one digits.

For example, the number 35461298 is written as follows:

This number is broken down into classes:

482 - first class (unit class)

630 - second class (class of thousands)

35 - third class (class of millions)

Discharge

Each of the numbers that make up the class is called its category, which also counts down to the right.

For example, the number 35 630 482 can be decomposed into classes and categories:

482 - first class

2 - first digit (units place)

8 - second place (tens place)

4 - third rank (hundreds rank)

630 - second class

0 - first digit (thousand units)

3 - the second category (tens of thousands)

6 - third rank (hundreds of thousands rank)

35 - third grade

5 - the first digit (place of million units)

3 - the second category (tens of millions)

The number 35 630 482 reads:

Thirty-five million six hundred thirty thousand four hundred eighty-two.

Rational counting problems and how to fix them

Rational memorization techniques.

As a result of questionnaires and observations from the lessons, I noticed that some of the students poorly solve various problems and exercises because they are not familiar with rational calculation techniques.

1. One of the techniques is to bring the material under study into a system that is convenient for memorization and preservation in memory.

2. In order for the memorized material to be stored in memory in a certain system, it is necessary to do some work on its content.

3. Then you can start assimilating each separate part of the text, rereading it and trying to immediately reproduce (repeat to yourself or aloud) what you read.

4. Repetition of the material is of great importance for memorization. This is also evidenced by the popular proverb: "Repetition is the mother of learning." But it is also necessary to repeat it intelligently and correctly.

The work of repetition must be revived by drawing on illustrations or examples that were not there before or they have already been forgotten.

Based on the foregoing, we can briefly formulate the following recommendations for the successful assimilation of educational material:

1. Set a task, quickly and firmly memorize educational material for a long time.

2. Focus on what needs to be learned.

3. Understand the teaching material well.

4. Make a plan of the memorized text, highlighting the main ideas in it, break the text into parts.

5. If the material is large, sequentially assimilate one part after another, and then present everything as a whole.

6. After reading the material, you need to reproduce it (tell what you read).

7. Repeat the material before it is forgotten.

8. Spread the repetition for a longer time.

9. When memorizing, use different types of memory (primarily semantic) and some individual features of your memory (visual, auditory or motor).

10. Difficult material should be repeated before going to bed, and then in the morning, "for fresh memory."

11. Try to apply the knowledge gained in practice. This is the best way to preserve them in memory (it is not for nothing that they say: "The real mother of learning is not repetition, but application").

12. It is necessary to acquire more knowledge, to learn something new.

Now you have learned how to quickly and correctly memorize the studied material.

An interesting trick of multiplying some numbers by 9 in combination with the addition of consecutive natural numbers from 2 to 10

12345x9 + 6 \u003d 111111

123456x9 + 7 \u003d 1111111

1234567x9 + 8 \u003d 11111111

12345678x9 + 9 \u003d 111111111

123456789x9 + 10 \u003d 1111111111

Interesting game "Guess the number"

Have you played the Guess the Number game? This is a very simple game. Let's say I guess a natural number less than 100, write it down on paper (so that there is no opportunity to cheat), and you try to guess it by asking questions that can only be answered "yes" or "no." Then you guess the number, and I try to guess it. Whoever guesses with fewer questions won.

How many questions do you need to guess my number? Do not know? I am going to guess your number by asking just seven questions. How? And here, for example, how. Let you guess the number. I ask, "Is it less than 64?" - "Yes". - "Less than 32?" - "Yes". - "Less than 16?" - "Yes". - "Less than 8?" - "Not". - "Less than 12?" - "Not". - "Less than 14?" - "Yes". - "Less than 13?" - "Not". - "The number 13 is conceived."

Clear? I divide the set of possible numbers in half, then the remaining half in half again, and so on, until there is one number in the remaining half.

If you liked the game or, on the contrary, you want more, then go to the library and take the book “A. P. Savin (Mathematical miniatures). In this book you will find a lot of interesting and exciting things. Book Image:

Thank you all for your attention

And I wish you success !!!

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What is the secret to rational counting?

Purpose of work: search for information, study of existing methods and techniques of rational calculation, their application in practice.

tasks: 1. Conduct a mini-research in the form of a survey among parallel classes. 2. Analyze on the topic of research: literature available in the school library, information in the academic textbook on mathematics for grade 5, as well as on the Internet. 3. Choose the most effective methods and means of rational counting. 4. Conduct a classification of existing techniques for rapid oral and written counting. 5. Create a Memo containing the techniques of rational counting for use in parallel 5 classes.

As I said, the topic of rational counting is relevant not only to students, but also to every person, in order to be convinced of this, I conducted a poll among students in grade 5. Questions and answers of the survey are presented to you in the application.

What is a rational account? A rational account is a convenient account (the word rational means convenient, correct)

Why do students have difficulties ???

Here are some assumptions: The student: 1. has a poor understanding of the topic studied; 2. does not repeat the material; 3. has poor numeracy skills; 4 . believes that it will not be useful to him.

Rational methods of oral and written calculations. In work and everyday life, the need for various kinds of calculations constantly arises. Using the simplest methods of verbal counting reduces fatigue, develops attention and memory.

There are four known methods of addition to speed up the calculations. I. Techniques for simplified addition of numbers

The method of sequential bitwise addition is used in oral calculations, as it simplifies and speeds up the addition of terms. When using this method, addition begins with the highest digits: the corresponding digits of the second term are added to the first term. Example. Find the sum of the numbers 5287 and 3564 using this method. Decision. We will make the calculation in the following sequence: 5 287 + 3 000 \u003d 8 287; 8 287 + 500 \u003d 8 787; 8 787 + 60 \u003d 8 847; 8 847 + 4 \u003d 8 851. Answer: 8 851.

Another method of sequential bitwise addition is that the highest bit of the second term is added to the highest bit of the first term, then the next bit of the second term is added to the next bit of the first term, etc. Let's consider this variant of the solution on the given example, we get: 5,000 + 3,000 \u003d 8,000; 200 + 500 \u003d 700; 80 + 60 \u003d 140; 7 + 4 \u003d 11 Answer: 8851.

Round number method. A number ending in one or more zeros is called a round number. This method is used when one can choose from two or more terms that can be completed to a round number. The difference between the round number and the number specified in the calculation condition is called the complement. For example, 1,000 - 978 \u003d 22. In this case, 22 is the complement of 978 to 1,000. To make addition in the way of a round number, it is necessary to round one or several terms close to round numbers, add round numbers and subtract arithmetic additions from the resulting sum. Example. Find the sum of the numbers 1 238 and 193 using the round number method. Decision. Let's round the number 193 to 200 and add as follows: 1 238 + 193 \u003d (1 238 + 200) - 7 \u003d 1 431.

A way of grouping the terms. This method is used in the case when the terms, when grouped together, add up to round numbers, which are then added together. Example. Find the sum of the numbers 74, 32, 67, 48, 33 and 26. Solution. Let's sum up the numbers grouped as follows: (74 + 26) + (32 + 48) + (67 + 33) \u003d 280.

Addition method based on the grouping of terms. Example: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + ……. + 97 + 98 + 99 + 100 \u003d (1 + 100) + (2 + 99) + (3 + 98) \u003d 101x50 \u003d 5050.

II. Simplified number subtraction techniques

A method of sequential bitwise subtraction. In this way, a sequential subtraction of each digit is subtracted from the decremented one. It is used when numbers cannot be rounded. Example. Find the difference between the numbers 721 and 398. Let's perform the steps to find the difference of the given numbers in the following sequence: represent the number 398 as a sum: 300 + 90 + 8 \u003d 398; let's perform a bitwise subtraction: 721 - 300 \u003d 421; 421 - 90 \u003d 331; 331 - 8 \u003d 323.

Round number method. This method is used when the subtracted is close to a round number. For the calculation, it is necessary to subtract the subtracted, taken as a round number, from the reduced, and add the arithmetic addition to the resulting difference. Example. Let's calculate the difference of numbers 235 and 197 using the round number method. Decision. 235 - 197 \u003d 235 - 200 + 3 \u003d 38.

III. Simple multiplication techniques

Multiplication by one followed by zeros. When a number is multiplied by a number that includes one followed by zeros (10; 100; 1,000, etc.), as many zeros are assigned to it on the right as there are in the factor after one. Example. Find the product of the numbers 568 and 100. Solution. 568 x 100 \u003d 56 800.

Method of sequential bitwise multiplication. This method is used when multiplying a number by any single digit. If you need to multiply a two-digit (three-, four-digit, etc.) number by a single-digit number, then first one of the factors is multiplied by tens of another factor, then by its units and the resulting products are summed up. Example. Find the product of the numbers 39 and 7. Decision. 39 x 7 \u003d (30 x 7) + (9 x 7) \u003d 210 + 63 \u003d 273.

Round number method. Use this method only when one of the factors is close to a round number. The multiplier is multiplied by a round number, and then by the arithmetic addition, and at the end the second is subtracted from the first product. Example. Find the product of the numbers 174 and 69. Decision. 174 x 69 \u003d (174 x 70) - (174 x 1) \u003d 12 180 - 174 \u003d 12 006.

A way to decompose one of the factors. In this method, one of the factors is first decomposed into parts (terms), then the second factor is alternately multiplied by each part of the first factor and the resulting products are summed up. Example. Find the product of the numbers 13 and 325. Decision. Let us expand the number into terms: 13 \u003d 10 + 3. Multiply each of the resulting terms by 325: 10 x 325 \u003d 3 250; 3 x 325 \u003d 975 Summarize the resulting products: 3 250 + 975 \u003d 4 225.

Secrets of quick verbal counting. Oral counting systems exist that allow you to count orally quickly and efficiently. We will look at some of the most commonly used techniques.

Multiply a two-digit number by 11.

Examples: 23x11 \u003d 23x (10 + 1) \u003d 23x10 + 23x1 \u003d 253 (distribution law of multiplication relative to addition) 23x11 \u003d (20 + 3) х 11 \u003d 20x11 + 3x11 \u003d 253 (distribution law and round number method) We studied this method , but we did not know another secret of multiplying two-digit numbers by 11.

Observing the results obtained when multiplying two-digit numbers by 11, I noticed that you can get the answer in a more convenient way: when multiplying a two-digit number by 11, the digits are moved apart and the sum of these digits is put in the middle. Examples. a) 23 11 \u003d 253, since 2 + 3 \u003d 5; b) 45 11 \u003d 495, since 4 + 5 \u003d 9; c) 57 11 \u003d 627, because 5 + 7 \u003d 12, two were placed in the middle, and one was added to the category of hundreds; I found confirmation of this method on the Internet.

2) The product of two-digit numbers, which have the same number of tens, and the sum of the ones is 10, that is, 23 27; 34 36; 52 58, etc. Rule: the figure of tens is multiplied by the next figure in the natural series, the result is recorded and the product of units is attributed to it. Examples. a) 23 27 \u003d 621. How did you get 621? We multiply the number 2 by 3 (the “two” is followed by the “three”), it will be 6, and next to it we will add the product of ones: 3 7 \u003d 21, it turns out 621. b) 34 36 \u003d 1224, since 3 4 \u003d 12, we assign 24 to the number 12, this is the product of units of these numbers: 4 6.

3) Division of three-digit numbers, consisting of the same digits, by the number 37. The result is equal to the sum of these identical digits of a three-digit number (or a number equal to three times the digit of a three-digit number). Examples. a) 222: 37 \u003d 6. This is the sum of 2 + 2 + 2 \u003d 6. b) 333: 37 \u003d 9, since 3 + 3 + 3 \u003d 9. c) 777: 37 \u003d 21, i.e. 7 + 7 + 7 \u003d 21. d) 888: 37 \u003d 24, since 8 + 8 + 8 \u003d 24. We also take into account that 888: 24 \u003d 37.

Learning the skills of rational oral counting will make your work more efficient. This is possible only with a good mastery of all the above arithmetic operations. The use of rational counting techniques speeds up calculations and provides the required accuracy.

Conclusion To unravel the main secret in the topic of my work, I had to work hard - to search, analyze information, question classmates, repeat the early known methods and find many unfamiliar ways of rational counting, and, finally, understand what is its secret? And I realized that the main thing is to know and be able to apply the known, find new rational methods of counting, know the multiplication table, the composition of the number (classes and categories), the laws of arithmetic operations. In addition, look for new ways to:

Techniques for simplified addition of numbers: (method of sequential bitwise addition; method of a round number; method of decomposing one of the factors into terms); - Techniques for simplified subtraction of numbers (method of sequential bitwise subtraction; method of a round number); - Techniques for simplified multiplication of numbers (multiplication by one followed by zeros; method of sequential bitwise multiplication; method of a round number; method of decomposing one of the factors; - Secrets of rapid oral counting (multiplying a two-digit number by 11: when multiplying a two-digit number by 11, the digits of this number are moved apart and in the middle they put the sum of these digits; the product of two-digit numbers, which have the same number of tens, and the sum of ones is 10; Division of three-digit numbers, consisting of the same digits, by the number 37. Probably, there are many more such ways, so I will continue to work over this topic next year.

In conclusion, I would like to end my speech with the following words:

Thank you all for your attention, I wish you success !!!