Ege in mathematics profile level

The work consists of 19 tasks.
Part 1:
8 tasks with a brief response of the basic level of complexity.
Part 2:
4 tasks with a brief answer
7 tasks with a detailed answer high level difficulties.

Performance time - 3 hours 55 minutes.

Examples of the tasks of the EGE

Solving the tasks of the exam in mathematics.

For self solutions:

1 kilowatt hour electricity costs 1 ruble 80 kopecks.
Electricity meter on November 1 showed 12625 kilowatt-hours, and on December 1 showed 12802 kilowatt-hour.
What amount should you pay for electricity for November?
Give the answer in rubles.

In the exchange paragraph 1 of the hryvnia costs 3 rubles 70 kopecks.
Vacationers exchanged rubles to the hryvnia and bought 3 kg of tomatoes at a price of 4 hryvnia per 1 kg.
How many rubles did this purchase cost them? Answer round up to an integer.

Masha sent SMS with New Year's Eve greetings to her 16 friends.
The cost of one SMS message 1 ruble 30 kopecks. Before sending a message to the account, Masha had 30 rubles.
How many rubles will remain from Masha after sending all messages?

The school has triple tourist tents.
What the smallest number Tents need to take a hike, in which 20 people participate?

The train Novosibirsk-Krasnoyarsk departs at 15:20, and arrives at 4:20 the next day (Moscow time).
How many hours the train is on the way?

Solve the equation:

1 / COS 2 x + 3TGX - 5 \u003d 0

Specify the roots
Segment belonging (-P; P / 2).

Decision:

1) We write the equation so:

(TG 2 x +1) + 3TGX - 5 \u003d 0

TG 2 x + 3TGX - 4 \u003d 0

tGX \u003d 1 or TGX \u003d -4.

Hence:

X \u003d n / 4 + pk or x \u003d -arctg4 + pk.

Segment (-P; P / 2)

Owned roots -3p / 4, -Arctg4, p / 4.

Answer: -3p / 4, -ArCTG4, P / 4.

Do you know what?

If you multiply your age by 7, then multiply by 1443, then the result will be your age written three times in a row.

We consider negative numbers with something natural, but it was not always. For the first time, the negative numbers were legalized in China in the III century, but were used only for exceptional cases, as they were considered, in general, framed. A little later, negative numbers began to be used in India to designate debts, but we did not fit the west - the famous Diofant Alexandria argued that equation is 4x + 20 \u003d 0 - absurd.

American mathematician George Dzig, being a postgraduate student at the university, once late for a lesson and accepted the equation written on the board homework. It seemed more difficult to him as usual, but after a few days he was able to execute it. It turned out that he decided two "unresolved" problems in statistics, which many scientists were fighting.

In Russian mathematical literature, Zero is not a natural number, and in Western, on the contrary, belongs to a variety of natural numbers.

Used by us decimal system The number arose due to the fact that the person in the hands of 10 fingers. The ability to an abstract account appeared in people not immediately, but it was the most convenient to use for the score. Mayan civilization and regardless of them Chukchi historically used a twenty number system, applying fingers not only hands, but also legs. At the heart of the twelve and sixty-digit systems common in the ancient Suchmera and Babylon, the use of hands was also: the phalanxes of other fingers of the palm were counted with a thumb, the number of which is 12.

One familiar lady requested Einstein to call her, but warned that her phone number is very difficult to remember: - 24-361. Remember? Repeat! Surprised Einstein replied: - Of course, I remembered! Two dozen and 19 square.

Stephen Hawking is one of the largest physicists of theorists and a popularizer of science. In a story about himself, Hoking mentioned that he became a professor of mathematics, not receiving any mathematical education since the secondary school. When Hawking began teaching mathematics in Oxford, he read a tutorial, ahead of his own students for two weeks.

The maximum number that can be recorded by Roman numerals, without breaking the rules of Schwartzman (Roman digits) - 3999 (MMMCMXCIX) - more than three digits in a row can not write.

It is known a lot of parables about how one person offers another pay with him for some service as follows: on the first cell of the chessboard, he will put one rice grain, on the second - two and so on: for each next cell twice as much as the previous one. As a result, the one who pays in this way will certainly ruin. This is not surprising: it is estimated that general weight Rice will make more than 460 billion tons.

In many sources, it is often in order to encourage poorly spending students, the approval is found that Einstein wounds in school mathematics or, moreover, he studied out of the hands badly in all subjects. In fact, everything was not true: Albert began to be talent in mathematics at an early age and knew her far beyond the school program.

Ege 2020 in mathematics task 19 with decision

Demonstration options 2020 in mathematics

EGE in mathematics 2020 in PDF format Basic level | Profile level

Tasks for preparation for the exam in mathematics: basic and profile level with answers and solutions.

Mathematics: Basic | Profile 1-12 | | | | | | | | the main

Ege 2020 in mathematics Task 19

EGE 2020 in mathematics profile level reference 19 with decision

Ege in mathematics

The number p is equal to the product of 11 different natural numbers, large 1.
What the smallest number of natural divisors (including the unit and the number itself) may have the number P.

Any natural number N represents the work:

N \u003d (p1 x k1) (p2 x k2) ... etc

Where P1, P2, etc. - Simple numbers,

A K1, K2, etc. - whole non-negative numbers.

For example:

15 = (3 1) (5 1)

72 \u003d 8 x 9 \u003d (2 x 3) (3 2)

So, the total number of natural divisors N is equal

(K1 + 1) (K2 + 1) ...

So, by condition, p \u003d n1 n2 ... n11, where
N1 \u003d (p1 x k) (p2 x k) ...
N2 \u003d (p1 x k) (p2 x k) ...
...,
And this means that
P \u003d (p1 x (k + k + ... + k)) (p2 x (k + k + ... + k)) ...

And the total number of natural dividers of the number p is equal

(k + k + ... + k + 1) (k + k + ... + k + 1) ...

This expression takes the minimum value if all numbers N1 ... N11 are sequential natural degrees of the same simple number, starting from 1: n1 \u003d p, n2 \u003d p 2, ... n11 \u003d p 1 1.

That is, for example,
N1 \u003d 2 1 \u003d 2,
N2 \u003d 2 2 \u003d 4,
N3 \u003d 2 3 \u003d 8,
...
N11 \u003d 2 1 1 \u003d 2048.

Then the number of natural divisors of the number P is equal
1 + (1 + 2 + 3 + ... + 11) = 67.

Ege in mathematics

Find all natural numbers,
Not representable in the form of the sum of two mutually simple numbers other than 1.

Decision:

Each natural number can be either even (2 K) or odd (2 k + 1).

1. If the number is odd:
n \u003d 2 k + 1 \u003d (k) + (k + 1). Numbers k and k + 1 always mutually simple

(If there is some number D, which is a divider x and y, then the number | XY | also must be divided into d. (k + 1) - (k) \u003d 1, that is, 1 should be divided into D, that is, D \u003d 1, And this is the proof of mutual simplicity)

That is, we have proven that all odd numbers can be represented as the sum of two mutually simple.
The exception by condition will be the number 1 and 3, since 1 can not be submitted in the form of the sum of natural, and 3 \u003d 2 + 1 and nothing else, and the unit as the foundation is not suitable under the condition.

2. If the number is even:
n \u003d 2 k
Here you have to consider two cases:

2.1. k - even, i.e. Representative in the form K \u003d 2 m.
Then n \u003d 4 m \u003d (2 m + 1) + (2 m-1).
Numbers (2 m + 1) and (2 m-1) may have a common divider only (see above), to which the number (2 m + 1) is divided - (2 m - 1) \u003d 2. 2 is divided by 1 and 2.
But if the divider is 2, it turns out that an odd number of 2 M + 1 must be divided by 2. This cannot be, therefore it remains only 1.

So we proved that all numbers of the form 4 m (that is, multiple 4) can also be represented as the sum of two mutually simple.
There is an exception - the number 4 (m \u003d 1), which, although it can be represented in the form of 1 + 3, but the unit as the foundation is still not suitable for us.

2.1. k is odd, i.e. Representative in the form K \u003d 2 M-1.
Then n \u003d 2 (2 m - 1) \u003d 4 m-2 \u003d (2 m-3) + (2 m + 1)
Numbers (2 m-3) and (2 m + 1) may have a common divider to which the number 4. That is, either 1, or 2, or 4. But neither 2, nor 4 are suitable, since (2 m + 1) - the number is odd, and no 2 cannot be divided into nor.

So we proved that all numbers of the form 4 M-2 (that is, all multiple 2, but not multiple 4) can also be represented as the sum of two mutually simple.
There are exceptions - numbers 2 (m \u003d 1) and 6 (m \u003d 2), which are one of the terms in the decomposition to a couple of mutually simple equal to one.

Give an example of a three-digit number, the amount of numbers of which is 20, and the sum of the squares of the numbers is divided into 3, but not divisible by 9.

Decision.

Spatize the number 20 to the well-known ways:

20 = 9 + 9 + 2 = 9 + 8 + 3 = 9 + 7 + 4 = 9 + 6 + 5 = 8 + 8 + 4 = 8 + 7 + 5 = 8 + 6 + 6 = 7 + 7 + 6.

With the decomposition of methods 1-4, 7 and 8, the numbers of the numbers are not more than three. When decomposition in the fifth way, the sum of the squares of a multiple nine. Decomposition by the sixth way satisfies the conditions of the task. Thus, the condition of the problem satisfies any number recorded by figures 5, 7 and 8, for example, the number 578.

Answer: 578 | 587 | 758 | 785 | 857 | 875

Source: demo version of the exam - 2015.

Find a three-digit natural number, more than 400, which, when divided by 6 and 5, gives equal non-zero residues and the first on the left of the number of which is the average arithmetic two other digits. In response, specify any one such number.

Decision.

The number has the same residues during division by 5 and 6, therefore, the number has the same residue during division by 30, and this residue is not zero and less than five. Thus, the desired number may be of the form :.

At. None of the numbers are not more than 400

With: 421, 422, 423, 424. The first left number is not an average arithmetic two other digits.

At: 451, 452, 453, 454. The number 453 satisfies all the conditions of the task.

Also suitable numbers 573 and 693.

Answer: 453,573, 693.

Answer: 453 | 573 | 693

Find a four-digit number, multiple 22, the product of the numbers of which is 24. In response, specify any one such number.

Decision.

In order for the number of ABCD to be divided into 22, it should be divided into 2, and 11. The product of numbers 24 can be represented by many ways, the basis of which are works -. Sign of divisibility by 11: The number is divided into 11 if the sum of numbers that stand on even places is equal to the amount of numbers on odd places, or differs from it by 11. Thus, A + C \u003d B + D or A + C \u003d B + D + 11 or A + C + 11 \u003d B + D. In addition, the number is divided by 2, then it must be even. According to the listed features, you can choose the following numbers: 4312, 2134, 1342, 3124

Awn: 2134 | 4312 | 1342 | 3124

Find a three-digit number, multiple 25, all the numbers of which are different, and the sum of the squares of the numbers is divided into 3, but it is not divided into 9. In response, specify any one such number.

Decision.

So that the number is divided by 25, it should end up 00, 25, 50 or 75. Our number can not be completed, since all its numbers must be different. We drink all three-digit numbers ending with 25, 50 or 75, all the numbers of which are different, will find the sum of the squares of their numbers, check if it is divided by 3 and by 9.

The amount of numbers is not divided into 3.

The amount of numbers is divided into 3, but not divisible to 9. This is the desired number.

The amount of numbers is not divided into 3.

The amount of numbers is divided into 3, but not divisible to 9. This is the desired number.

The amount of numbers is not divided into 3.

The amount of numbers is not divided into 3.

The amount of numbers is not divided into 3.

The amount of numbers is not divided into 3.

The amount of numbers is divided by 3 and by 9.

The amount of numbers is not divided into 3.

The amount of numbers is not divided into 3.

The amount of numbers is not divided into 3.

The amount of numbers is divided into 3, but not divisible to 9. This is the desired number.

The amount of numbers is not divided into 3.

The amount of numbers is divided into 3, but not divisible to 9. This is the desired number.

The amount of numbers is not divided into 3.

The amount of numbers is divided into 3, but not divisible to 9. This is the desired number.

The amount of numbers is not divided into 3.

The amount of numbers is not divided into 3.

The amount of numbers is not divided into 3.

Average general education

Merzlyak Line. Algebra and start analysis (10-11) (y)

Line UMK A. G. Merzlyak. Algebra and the start of analysis (10-11) (b)

Line Ukk G. K. Moravina. Algebra and the beginning of mathematical analysis (10-11) (coal.)

Line UMK G.K. Muravina, K.S. Maravina, O.V. Vigorous. Algebra and began mathematical analysis (10-11) (bases)

EGE-2018 in mathematics, basic level: Task 19

We offer to your attention 19 the tasks of the ege 2018 in mathematics. The article contains detailed analysis Tasks, algorithm of solutions and recommendations of topical manuals for preparation for the EEG, as well as a selection of materials in mathematics published earlier.

Mathematics: algebra and began mathematical analysis, geometry. Algebra and beginning of mathematical analysis. Grade 11. A basic level of

The textbook is included in the CMD in mathematics for the 10-11 classes studying the subject on basic level. The theoretical material is divided into mandatory and additional, the system of tasks is differentiated by the level of complexity, each chapter item is completed by control issues and tasks, and each chapter - home control work. The textbook includes project topics and made links to Internet resources.

Task 19.

More than 40, but less than 48 integers are written on the board. The arithmetic average of these numbers is -3, the arithmetic average of all positives is 4, and the arithmetic average of all negative is -8 equal.

a) How many numbers are written on the board?

b) What numbers are written more: positive or negative?

in which the greatest number Positive numbers can be among them?

Decision

A) Let them among the written numbers

x. - Positive

y. - Negative

z. - zerule

Then we have that

• the amount of positive numbers is equal to 4 x.
• the sum of the negative numbers is -8 y.
• the sum of all numbers of the series 4 x. + (–8y.) + 0z. = –3(x. + y. + z.)

4(x. – 2y. + 0z.) = –3(x. + y. + z.)

Because The left part of the equality of Paint 4, the right part of the equality should be more than 4, which means

x. + y. + z.(Number of numbers) multiple 4.

40 < X. + y. + z.< 48,

x. + y. + z.= 44

So on the board written 44 numbers.

B) Consider equality 4 x. + (–8y.) + 0z. = –3(x. + y. + z.)

4x.– 8y.= – 3x.– 3y.– 3z.

4x. + 3x. + 3z. = 8y. – 3y.

7x. + 3z. = 5y.

From here we get, because z ≥ 0 (number of zeros in row)

7x. < 5y.

x. < y.

So positive numbers are less than negative.

C) because x. + y. + z. \u003d 44, we will substitute this value in equality 4 x.+ (–8y.) + 0z. = –3(x. + y. + z.),

4x.– 8y. \u003d (-3 · 44) / 4

x -2y. = –33

x. = 2y. – 33

Considering that x. + y. + z. \u003d 44, we have x. + y. ≤ 44, substitute x. = 2y. - 33 in this inequality

2y. – 33 +y.≤ 44

3y. ≤ 77

y.≤ 25	2
	3

y.≤ 25, given that x. = 2y. - 33 Receive x. ≤ 17.

Task №19 from the basic exam in mathematicsMathVideourok.moy.su

Signs of divisibility on 2 and 4:

The number is divided into 2 if it ends even
Figure or zero.
Numbers 2346 and 3650 - divided by 2. The number 4521 is not
It is divided into 2.
The number is divided into 4 if the two last
null numbers or form a number divided by 4. in

Numbers 31700 and 16608 - on 4. 215634 - not
It is divided into 4.

Signs of divisibility on 3 and 9:

Only only those numbers in which the amount
The numbers are divided into 3.
Numbers 17835 and 5472 - divided by 3. Number 105499 - not
It is divided into 3.
Only ones are only those numbers that
The numbers are divided into 9.
Numbers 2376 and 342000 - divided by 9. Number 106499 - not
It is divided into 9.

Signs of divisibility on 8 and 6:

The number is divided into 8 if the last three numbers of it
zeros or form a number divided by 8. In
Other cases - not divisible.
Numbers 125000 and 111120 - divided by 8. Numbers 170004 and
124300 - not divided by 8.
The number is divided by 6 if it is divided simultaneously
2 and by 3. Otherwise, it is not divisible.
Numbers 126 and 254610 - divided by 6. The numbers 3585 and 6574 are not divided by 6.

Signs of divisibility on 5 and 25:

5 are divided by numbers, the last figure of which 0
or 5. Others - do not share.
Numbers 245 and 56780 - divided by 5. Numbers 451 and 678 - not
divided by 5.
25 are divided by numbers, the last two numbers of which
zeros or form a number divided by 25 (i.e.
numbers ending at 00, 25, 50 or 75). Others
Do not share.
Numbers 7150 and 345600 - divided by 25. Number 56755 - not
It is divided by 25.

Signs of divisibility at 10, 100 and 1000:

Only those numbers, the last figure divide on 10
which zero, per 100 - only those numbers that
The last two digits of zeros, per 1000 - only those
which are three last digits of zeros.
The number 34680 - divided by 10. The number 56700 - divides on
100 and 10. The number 87549000 is divided by 10, 100 and 1000.
Numbers 75864, 7776539 and 9864032 - do not divide 10, 100 and
1000.

Sign of divisibility at 11:

Only there are only those numbers that the amount of numbers,
occupying odd places or equal to the amount of numbers,
occupying even places or varies from it by number,
divided by 11.
The number 103785 is divided into 11, as the amount of numbers occupying
odd places, 1 + 3 + 8 \u003d 12 is equal to the amount of numbers that occupy even
Places 0 + 7 + 5 \u003d 12.
The number 9163627 is divided into 11, as the amount of numbers occupying
odd places, there are 9 + 6 + 6 + 7 \u003d 28, and the amount of numbers occupying
even places, there is 1 + 3 +2 \u003d 6; The difference between numbers 28 and 6 is
22, and this number is divided by 11.
The number 461025 is not divided into 11, since the numbers 4+ 1 + 2 \u003d 7 and b +0 +
5 \u003d 11 are not equal to each other, and their difference 11 -7 \u003d 4 is not divided into 11.

To begin with, consider an example - solution of the problem 19. (on this topic integers ) - Kim Real Ege 2015. years, early period, basic level. (Theory to her - signs of divisibility - below.)

Task 19.

Disture 181615121 three digits so that the resulting number is divided by 12. In response, specify any one such number.

Decision.

We declare the divider - the number 12 on simple factors. 12 \u003d 3 × 4 \u003d 3 × 2 × 2.
Therefore, the specified number after crossing numbers should be divided into 3 and 4 or 2, once again at 2 and, finally, by 3.
On 2, there are even numbers, therefore 1 at the end strike at once. It will remain 18161512.
But we need it to share 2 twice, i.e. shared on 4.
A sign of divisibility on 4 argues that for this, 4 should be divided into a two-digit number formed by the latest two-digit. 12 : 4 \u003d 3, so the two last numbers of the number 18161512 can not be deleted. They guarantee the division of a number of 4 (on both twos).
So that the number is shared by 3, it is necessary that the sum of its numbers shared on 3.
1+8+1+6+1+5+1+2=25
25 \u003d 3 × 8 + 1 - You can delete one of the units, but by the task condition you need to strike two more numbers;
25 \u003d 3 × 7 + 4 - no two digits for deletion, the sum of which would be 4, because The last figures 1 and 2 cannot be touched;
25 \u003d 3 × 6 + 7 - the sum of the two delicted numbers will be 7, if you draw 6-ku and any of the units other than the last one.
So, possible answers: 811512 or 181512. We choose one of them, for example

Answer: 181512.

Comment: On the real exam, check your answer to the division in the column.

Someone may have questions that such simple factors and how to put on simple factors?
Simple factors can not be further divided. Simple numbers are divided only on themselves and 1, for example, 13: 1 \u003d 13 or 13:13 \u003d 1 and that's it. And to lay it better gradually.
For example, 60 \u003d 6 × 10, 6 \u003d 2 × 3 and 10 \u003d 2 × 5, it means 60 \u003d 2 × 3 × 2 × 5.

To solve such tasks, you need to know theorems - signs of the divisibility of natural numbers. The more you know the signs, the faster you decide the task. Repeat the main ones.

Signs of the divisibility of natural numbers

Since humanity has invented ordinary and decimal fractions, we can apply the division operation to any values. However, the concept dividitude of numbers Usually considered on the set of natural numbers. When we say "the number is divided", then we mean that the division occurs without a residue and the result of the division is also a natural number.

Sign of divisibility by 2.

On 2 divided by all other numbers. We are because we call them younger.

The number is divided into two if and only if his last digit is divided into 2, i.e. 2, 4, 6, 8, 0.

Sign of divisibility by 3.

The natural number is divided into three if and only if the amount of his numbers is divided by 3.

For example, 4539861 is divided into 3, because 4 + 5 + 3 + 9 + 8 + 6 + 1 \u003d 36. The number 36 is divided into 3.
For example, 394762 is not divided into 3, because 3 + 9 + 4 + 7 + 6 + 2 \u003d 31. The number 31 is not divided into 3.
You can check with your favorite calculator
4539861: 3=1513287
394762: 3=131587,33333333333333333333333333

If the amount of numbers turned out to be a multivalued number, its divisibility can be checked by the same feature.
For example, 16539478617177984079 is divided into 3, because 1 + 6 + 5 + 3 + 9 + 4 + 7 + 8 + 6 + 1 + 7 + 1 + 2 + 7 + 7 + 9 + 8 + 4 + 0 + 7 + 9 \u003d 111. 111 divided by 3, because 1 + 1 + 1 \u003d 3. The number 3 is divided into 3.
165394786171277984079: 3 = 55131595390425994693

Sign of divisibility by 4.

A natural number containing at least three digits is divided into 4 if and only if it is divided into 4 two-digit number formed by the last two digits of a given number.

As for checking the divisibility by 4 double digits, we use the fact that 4 \u003d 2 × 2, i.e. Divide on 4 - the same thing that is twice in a row to divide on 2. Therefore, firstly, the two-digit number should be even, and, secondly, it is easy to divide on 2 and see whether the result is also even number. For example,

5773211789020783 is not divided into 4, because 83 is not divided into 2.
4920904953478666 is not divided into 4, because 66. : 2 \u003d 33 - odd number.
5897592348940996 is divided into 4, because 96. : 2 \u003d 48 - a thorough number.

Proof of the performance of this feature is based on divisibility 100 on 4 and the amount of the divisibility theorem, which is shown below. Here we consider an explanation on the example from the given task of the USE.
18161512 \u003d 18161500 + 12 \u003d 181615 × 100 + 12 \u003d 181615 × 25 × 4 + 3 × 4 \u003d (181615 × 25 + 3) × 4.
In brackets, the natural number will be obtained, it means that the initial number can be divided into 4 without a residue.

Sign of divisibility by 5.

The number is divided by 5 if and only if its last digit is either 5 or 0.

Sign of divisibility on 6 It is usually not formulated as the theorem. Since 6 \u003d 2 × 3, then a sequentially used specimen is used by 2 and by 3. Thus, it is used for 6 parts, the amount of numbers of which is divided by 3.
629 - not divided by 6, odd.
692 - It is not divided into 6, which is, but 6 + 9 + 2 \u003d 17 is not divided into 3.
792 - It is divided into 6, which is also 7 + 9 + 2 \u003d 18 divided by 3.

Sign of divisibility on 8 It is also not formulated as the theorem.
Since 8 \u003d 2 × 4 and 1000 \u003d 250 × 4, therefore, for numbers more than 1000, by analogy with a sign of divisibility by 4, a division of 8 numbers formed by three last digits is checked, and for numbers less than 1000 (three-digit), sequentially divided into 2 and verify the result obtained on the basis of division by 4. For example,
58989081099472 - divided by 8, as 472 : 2 \u003d 236, and 36 divided by 4.

Sign of divisibility by 9.

The natural number is divided into 9 if and only if the amount of its numbers is divided into 9.

For example, 4539861 is divided into 9, because 4 + 5 + 3 + 9 + 8 + 6 + 1 \u003d 36. The number 36 is divided into 9.
For example, 394762 is not divided into 9, because 3 + 9 + 4 + 7 + 6 + 2 \u003d 31. The number 31 is not divided into 9.
4539861: 9=504429
394762: 9=43862,444444444444444444444444444

Sign of divisibility by 10.

The natural number is divided by 10 if and only if its last digit 0.

This feature is easy to spread to any degrees of dozens. The number is divided by 100 when the two of its last digits are zeros, per 1000, when at the end three zero, etc.

Easy memorable signs of divisibility on simple numbers of type 7, 11, 13, 17 ..., Unfortunately no. The EGE organizers know the tasks focused on the use of exclusively such solutions will not be included. Although for a long history of development of the technique of oral account, mathematics, of course, identified and formulated some common features of the divisibility of such numbers. Interested can refer to Wikipedia.

I would recommend only to pay attention to another 11. It is clear that the two-digit number is divided by 11 if it consists of identical numbers. The three-digit number is divided into 11 if its average digit is equal to the sum of two extreme, or if the sum of the first and last digits is equal to the average digit plus 11. For example, 495 is divided by 11, since 4 + 5 \u003d 9, and 957 is divided by 11, so As 9 + 7 \u003d 5 + 11.

And in memorization signs of divisibility for constituents not necessary. Composite numbers can be decomposed on simple multipliers.

Theorems on the divisibility of the work and the sum of natural numbers.

If in the work at least one of the factors is divided into some number, then composition It is divided into this number.

For example, a product of 475 × 1230 × 800 is divided into 3, since the second factor satisfies the sign of division by 3 - the sum of its numbers 1 + 2 + 3 + 0 \u003d 6 is divided by 3.

If each term is divided into a number, then sum It is divided into this number.

For example, the amount of 475 + 1230 + 800 is divided into 5, as each rogue satisfies the sign of division by 5.

The opposite statement of the division of the amount is not true. If each summary amount is not divided into some number, then for the amount both options are possible, as it is divided and it is not divided.
43 is not divided into 5, 17 is not divided by 5, 43 + 17 \u003d 60 divided by 5.

The opposite statement on the divisibility of the work can be formulated only after decomposition of the divisor to simple favors. Actually, this action was devoted to the task that was placed at the beginning of the section.

If you are friends with an algebra and know how to carry out a common factor for brackets and reduce ordinary fractions, then the theorem of the divisibility amount can be remembered as the presence of a common benchmark, and the theorem on the divisibility of the work, as an opportunity to reduce the ordinary fraction.

Using the amount of the amount of the amount, you can "save" on the calculations, for example, when checking the signs of divisibility by 3 and by 9. When you add large numbers, you can throw away all the numbers of obviously divided, respectively by 3 or 9.
Let's go back to the last example from the point "Sign of division by 3".
For the number 165394786171277984079 instead of 1 + 6 + 5 + 3 + 9 + 4 + 7 + 8 + 6 + 1 + 7 + 1 + 2 + 7 + 7 + 9 + 8 + 4 + 0 + 7 + 9 Calculate 1 + 5 + 4 + 7 + 8 + 1 + 7 + 1 + 2 + 7 + 7 + 8 + 4 + 0 + 7 \u003d 69. The result is the same - divided by 3.

And last:
Mathematics do not like to write a lot. Long suggestions and repetitions of the same words are good when explaining the solution, but it is advisable to use the symbols when it is desirable. For the term "divided" you can use a symbol ⋮ Vertical dot.
48⋮ 6 means that 48 is divided into 6, or that the number 48 is multiple of number 6.

Tasks for self-test.

Here are tasks with solutions that are temporarily hidden so that you can first think about them on your own, and then press the button to compare your own and my solutions. Similar tasks with checking your response can be found in the open bank of the tasks of the Federal Institute of Pedagogical Measurements.

Task 1.

Give an example of a five digit number of multiple 12, the product of the numbers of which is 40. In response, specify exactly one such number.

Show a decision

Spread the number 40 to simple multipliers. 40 \u003d 2 × 2 × 2 × 5.
There are only four such multipliers, the numbers are not enough for a five-digit number, but you can always add a unit into the work, the result will not change.
40 \u003d 2 × 2 × 2 × 5 × 1.
Thus, the number in response can be made only from these numbers: 1,2,2,2,5.
So that the number was multiple 12 (the same thing that was divided into 12 without the residue) it should satisfy the signs of divisibility by 3 and by 4, as 12 \u003d 3 × 4.
Check the amount of numbers 1 + 2 + 2 + 2 + 5 \u003d 12. It is divided by 3, so our number will be divided into 3 for any permutations of numbers.
And so that it shall be divided into 4, at the end you need to put two digits so that the number formed by them is divided by 4.
It is obvious that the last digit should be 2, others are odd. Check options 12, 22, 52.
12: 4 \u003d 3; 22: 4 \u003d 11: 2 - It is not divided by a lot; 52: 4 \u003d 13.
Conclusion: the number must be compiled so that at the end it was 12 or 52, and at the beginning, any permutations from the remaining three digits.
Possible answers: 12252, 21252, 22152, 22512, 25212, 52212. In response, we write one of them. For example,

Answer: 21252

Comment: Your decision should be somewhat shorter, because it is enough to find at least one of the possible answers.

Task 2.

Give an example of a three-digit number of multiple 15, the product of the numbers of which is 30. In response, specify exactly one such number.

Show a decision

Spread the number 30 to simple multipliers. 30 \u003d 2 × 3 × 5.
There are three such multipliers, we need to make a three-digit number, which is divided into 15, i.e. Satisfies signs of divisibility by 3 and by 5, since 15 \u003d 3 × 5.
So that the number is divided by 5, it should end the number 5.
Check the amount of numbers 2 + 3 + 5 \u003d 10. The amount of numbers is not divided into 3, so our number will not be divided into 3 for any permutations of numbers.
Dead end? Not. Repeater again, you can add any number of units as a factory and the result will not change.
Imagine 30 as 2 × 3 × 5 × 1.
Now possible digits for the preparation of a three-digit number more than needed. Therefore, we grouped some simple factors into the compound: 2 × 5 \u003d 10 and 3 × 5 \u003d 15 These are not numbers, but two-digit numbers. 2 × 3 \u003d 6 Number 6 is indicated by the number 6.
Imagine 30 as 6 × 5 × 1.
Check the amount of numbers 6 + 5 + 1 \u003d 12. It is divided into 3. Thus, the number in response can be made up from numbers: 6,51. The last digit should be 5.

Possible answers: 615, 165

Task 3.

The numbers of the four-digit number, multiple 5, recorded in the reverse order and received the second four-digit number. Then, from the first number, the second was detected and received 2277. Bring exactly one example of such a number.

Show a decision

The number, multiple 5, ends with numbers 0 or 5. Then the number recorded in the reverse order should begin with 0 or C. 5. If the number begins with 0, it will not be four-digit, and it will be three-digit, since 0 at the beginning is usually Do not write. For example, 0348 is just 348. So the desired number ends with a digit 5. The rest of its numbers will designate letters a, B, C. The number in this case is indicated aBC5____ .
The hell is needed here in order not to confuse this designation with the algebraic product of variables ( a. Multiply by b., multiply by from ...). The number recorded in the reverse order is indicated 5 cBA____ .
By condition

aBC5____ − 5cBA____ = 2277.
Imagine that we carry out this subtraction in the column.
1) 5 less than 7, then when subtracting had to occupy a dozen.
10 + 5 − a. = 7. a. = 15 − 7 = 8.
2) When subtracting dozens not so obviously, they occupied or did not occupy a unit in the discharge of hundreds. First, let's say that they did not occupy. Then from the number reduced per unit c. did you read b. and got 7.
(c. − 1) − b. = 7. c. = 8 + b..
This option is suitable b. \u003d 0 I. b. \u003d 1. Large values b. Enlarge c. up to a double digit. Avoid for example b. \u003d 1, then c. \u003d 9, and we are convinced that the number 8195 satisfies the condition of the problem.

Answer: 8195

Comment: Maybe another right answer 8085 if you choose b. \u003d 0 in step 2). Whether the assumption will work that when subtracting dozens occupied a unit in the discharge of hundreds, check it yourself.