Quest Source: Decision 3754. USE 2016. Mathematics, I. V. Yashchenko. 30 options for typical test items.
Task 19. On the board were written 20 natural numbers (not necessarily different), each of which does not exceed 40. Instead of some of the numbers (possibly one), they wrote numbers on the board that are less than the original ones by one. The numbers, which after that turned out to be equal to 0, were erased from the board.
a) Could it be that the arithmetic mean of the numbers on the board increased?
b) The arithmetic mean of the originally written numbers was equal to 27. Could the arithmetic mean of the numbers remaining on the board be equal to 34?
c) The arithmetic mean of the originally written numbers was 27. Find the largest possible value of the arithmetic mean of the numbers that remain on the board.
Solution.
a) Yes, maybe, for example, if you take 19 numbers equal to 10, and the 20 is equal to 1, then after decreasing the 20 number by 1, it becomes equal to 0 and the average value is no longer 20 numbers, but 19, then we have:
Initial mean:;
Average value after change: 
.
As you can see, the second average value has become larger than the initial one.
b) Suppose that to fulfill this condition, you need to take units, then take numbers and one number, a total of 20 numbers. Their arithmetic mean will be
,
and after erasing units should get
,
that is, we have a system of equations:
Subtracting the second from the first equation, we get:
Thus, to fulfill the condition of this paragraph, you need to take a fractional number of numbers, which is impossible within the framework of this task.
Answer: no.
v) To get the maximum average of the numbers remaining on the board, you first need to write down a set of numbers consisting of the largest number of ones (which, then, will be erased from the board), and the rest of the numbers must be maximum. We write this condition in the form
,
where is the number of units; - 20th number (it is chosen so as to provide an average of 27). Hence we have:
From the resulting expression, it can be seen that the minimum value at which we get the maximum value. Thus, we have a sequence of numbers, the sum of which is
nothing to do.
The task is a joke. Ira borrowed 100 rubles from her mother, but lost them. Then I borrowed 50 rubles from a friend. For 20 p. bought pies, and the remaining 30 rubles. returned to mom. It turns out that she owes her mother 70 rubles. plus 50 p. friend, only 120 rubles, plus 20 rubles, which I spent on pies. Total 140 rubles, but in total she must return 150 rubles. Question: where else is 10 rubles?
Solution. Ira lost and spent 100 + 20 = 120 rubles. And I must return exactly this money: to mother 100 - 30 = 70 rubles. and girlfriend 50 p. And all other calculations from the evil one.
1.7. Multiplication. The laws of multiplication
1.8. Distribution law
V Section 1.7 introduces the concept of a product of two numbers using the example of the product of numbers 3 and 4. Note that this product is the sum of three terms, each of which is equal to 4, that is, 3 ∙ 4 = 4 + 4 + 4. This is necessary so that further under 3 ∙ a understand the sum a + a + a. For any number a, the equality 1 ∙ a = a is considered true.
This approach to the definition of a work seems inconvenient, since in elementary school they say that 3 ∙ 4 is 3 + 3 + 3 + 3 (take 3 4 times). But this seeming inconvenience is eliminated in the very first lesson, as soon as it is shown that the displacement law of multiplication is valid.
The movement and combination laws of multiplication are explained when calculating the number of squares and the number of cubes.
For any number a, the equalities 0 ∙ a = 0, a ∙ 0 = 0 are considered true. In addition, the equality 0 ∙ 0 = 0 is true.
V Clause 1.8 explains the distribution law when calculating the number of squares, shows the application of the distribution law to open brackets and take the common factor out of brackets.
When studying all three laws, schoolchildren should be taught to write laws using letters denoting arbitrary numbers, and to memorize the formulations of laws. This helps the development of clear mathematical speech, gives
students "speech templates" for oral answers.
Here and below, students should be drawn to the advantages in the speed of computations that the one who possesses the learned laws has. Thus, the teacher creates an intra-subject motivation coming from the subject (and not from the outside) to learning.
RT. Using tasks 66–70 in the first lesson will allow you to repeat the multiplication table, draw the students' attention to the pairs of factors that give 10, 100, 1000, etc. in multiplication. Assignments 71–76 are aimed at practicing the application of the studied laws.
Decisions and comments
90. a) The number 12 was first increased by 2 times, the result obtained was increased by 3 times. What is the result?
b) We thought of a number, increased it 3 times, the result was increased 4 times. How many times did the number increase as a result?
Solution. a) 12 ∙ 2 = 24, 24 ∙ 3 = 72, the result is 72.
Here it is advisable to ask students: how many times has the number 12 increased in 2 times. The answer can be obtained using the combination law of multiplication: (12 ∙ 2) ∙ 3 = 12 ∙ (2 ∙ 3) = 12 ∙ 6 - the number 12 increased by 6 times in 2 times. This answer will prepare the students to solve the 90b problem on their own.
b) First, the problem can be solved for a specific conceived number, for example, 2 or 3. It turns out that in either case the conceived number has increased 12 times. To show that the answer in this problem does not really depend on the choice of the conceived number, we denote the conceived number by the letter a. Then (a ∙ 3) ∙ 4 = a ∙ (3 ∙ 4) = a ∙ 12 - the number a in 2 times increased by 12 times.
91. What laws are used in the following calculations:
20 ∙ 30 = (2 ∙ 10) ∙ (3 ∙ 10) = (2 ∙ 3) ∙ (10 ∙ 10) = 6 ∙ 100 = 600?
a) Calculate: 20 ∙ 50.
Solution. Both laws of multiplication were used: displacement and combination. Note that the above application of these laws is not shown in detail, for example, like this:
20 ∙ 30 = (2 ∙ 10) ∙ (3 ∙ 10) = ((2 ∙ 10) ∙ 3) ∙ 10 = (2 ∙ (10 ∙3)) ∙ 10 = = 2 ∙ (3 ∙ 10) ∙ 10 = ((2 ∙ 3) ∙ 10) ∙ 10 = (2 ∙ 3) ∙ (10 ∙ 10) = 6 ∙ 100 = 600,
since students do not yet have the motivation to be accurate in converting numeric expressions. However, when performing the following tasks, you may not require such an incomplete recording of solutions, which is given above. The solution can be written briefly: a) 20 ∙ 50 = 1000.
(27 + 73) = 356 100 + 644 100 = (356 + 644) 100 = 1000 100 =100 000.
Intermediate control. DM. C – 2.
1.9. Column addition and subtraction of numbers
1.10. Multiplication of numbers by a column
The purpose of these points is to demonstrate to students how the laws of addition and multiplication, the distribution law are used when adding, subtracting and multiplying multidigit numbers in a column. This does not imply that students should make similar justifications themselves, but it would be helpful for them to notice that the correctness of the columnar calculations follows from the correctness of the laws of addition and multiplication.
Particular attention should be paid to the correctness of signing multipliers under each other, the record of which ends with zeros.
From this point on, calculations in a column are included in the computational practice of fifth-graders, but it is necessary to draw the attention of students that sometimes calculations with multi-digit numbers can be easier to perform without a column if you notice pairs of numbers that give "round" sums (task 135); or if you notice that the common factor can be taken out of the brackets (task 144). It is necessary to develop and support in every possible way the desire of schoolchildren to calculate economically, and for this, as we have already noted, they are required to be observant
and possession of the studied theory.
V in the future, the desire to save time in calculations should become an incentive for the development of observation, as well as for
formation of the idea that knowledge of many theoretical information can simplify the solution of the problem.
RT. The use of assignments77, 78 in the first lesson on addition and subtraction in a column will intensify the learning process, since students only need to enter the answers in the columns already written down. Task79 prepares them for task80 and tasks133 and134 from the textbook. Task 81 is performed at the beginning of the study of column multiplication, while it is necessary to draw the students' attention to the recording of the multipliers. Task 82 is devoted to solving puzzles.
Decisions and comments
133. Correctly performed examples of addition and subtraction were written on the board, then some numbers were erased and replaced with letters. Rewrite the examples, replacing letters with numbers so that you get the correct entries again:
Hereinafter, students can receive answers by selecting a suitable number and checking the correctness of the answer, but it will be better if the board gives examples of reasoning: to get 8, add 5 to 3 (example "a"), etc.
Answer. a) 725 + 173 = 898; b) 952 - 664 = 288; c) 502 + 879 = 1381;
d) 1456 - 568 = 888.
134. Reconstruct the examples, assuming that the same letters mean the same numbers, and different letters mean different numbers:
linear search algorithm for the answer. At each step, he gives a single meaning to the letter.
1) The sum of two four-digit numbers is a five-digit number. Hence, d
1, i.e.
1raka 2) The sum p + p is a number ending in an even digit, i.e. a - even
number, but then (see the place of hundreds of the sum) a = 2, i.e.
1p2k2 3) The sum p + p is a number ending in 2, this is possible only in two
cases: p = 1 or p = 6. But the number 1 already exists (different letters correspond to different numbers), therefore, p = 6, i.e.
162k2 4) Thenak = 5, y = 8, i.e.
Example "c" has been restored, and all digits were found unambiguously.
d) This task is more complicated; when it is performed, a branching algorithm for finding an answer is implemented. At some step, it does not give the only meaning for the letter. The difficulty lies in remembering to complete the reasoning for each branch of the algorithm.
1) The sum of two six-digit numbers is a seven-digit number, therefore, and =
2) The sum b + b ends in an even digit, i.e. e is an even number, in the tens + l place is a number ending in an even digit. To get the number 1 in the tens place of the sum, it is necessary that it be ≥ 5 or = 0 or = 5.
3) If l = 0, i.e.
toa = 5, i.e. e. | |
1sde01e But then in the thousandth place the sums + m + 1 ends in an odd number, i.e. e
An odd number, but above it was established that e is an even number. The resulting contradiction means that 0. Hence, η = 5.
4) Since l = 5, i.e. e.
1sde51e then in the place of hundreds the sum a + a + 1 ends in 5. This is possible in two cases:
a = 2 or a = 7. But for a = 7 in the place of thousands, the number is odd, which is impossible, since it was established above that e is an even number. Therefore, a ≠ 7. Hence, a = 2.
5) Since a = 2, i.e.
1sde51e and since e is an even number, then it cannot be zero (if e = 0, then b = 0 or = 5,
which is impossible, since it has already been established that b ≥ 5, and the number 5 already exists). The number 2 already exists, therefore ≠ 2. Therefore, it remains to consider three possible cases: e = 4, e = 6,
e = 8.
6) If e = 4, then b = 7, then (see the thousand place) m = 2 or m = 7, which is impossible, since the numbers 2 and 7 are already there.
7) If e = 6, then in the place of tens of thousands of the sum d = 3 (since the number 2 is already
is), but then the sum will not be a seven-digit number, which is impossible. Hence, e = 8.
8) Since f = 8, then b = 9, m = 4, q = 6, s = 3, i.e. e.
Example "d" has been restored, and all the numbers have been found unambiguously. Showing solutions to examples "c" and "d" on a board is easier than publishing in
book, since in the case of a linear algorithm, using a rag and chalk, you can gradually replace letters with numbers and from this example with letters you can get the desired example with numbers. And in the case of a branching algorithm, it is necessary to leave on the board all not fully considered options. The scheme of the algorithm implemented when solving task d) can be depicted as follows:
Of course, students can just pick up the numbers they need, but then they won't be sure that the solution they have found is the only one.
135. a) Perform the steps: (5486 + 3578) + 1422.
Solution. Here I would like that, in addition to the ability to apply 2 times calculations in a column, one of the students noticed that the sum of the second and third numbers is "round", so the calculation can be easily performed in a line:
(5486 + 3578) + 1422 = 5486 + (3578 + 1422) = 5486 + 5000 = 10 486.
146. The product of four consecutive natural numbers is equal to
3024. Find these numbers.
Solution. Note that among the sought-for four numbers there is no number 10 and number 5, since if there were at least one of these factors, then the product would end in zero. It remains to check: 1 ∙ 2 ∙ 3 ∙ 4 = 24, 6 ∙ 7 ∙ 8 ∙ 9 = 3024.
Answer. 6, 7, 8, 9.
1.11. Degree with natural exponent
This clause introduces the concept of a degree with a natural exponent for cases n> 1 and n = 1. Students must master the terminology: degree, base of the degree (the number that we raise to a power), exponent (shows the degree to which we raise the base of the degree), square numbers, cube numbers, and learn how to calculate degrees.
RT. It is advisable to use tasks 83–86 at the initial stage of studying the material. When studying this item, you can use tasks 87–90.
Solutions and comments
171. Among the first five natural numbers, there are two unequal numbers m
and n such that n m = m n. Find these numbers.
Solution. These numbers are 2 and 4. Indeed, 24 = 2 ∙ 2 ∙ 2 ∙ 2 = 16, 42 = 4 ∙ 4 = = 16,
i.e. 24 = 42.
Answer. 2 and 4.
1.12. Whole division
This clause introduces the concept of whole division and the corresponding terminology, explains why any natural number or zero cannot be divided by zero. Examples of simplification of division in some cases are given. You should pay attention to the property of the quotient, which sometimes helps to simplify calculations (tasks 186-187). For example, when dividing a number by 5, you can dividend and
multiply the divisor by 2 and divide the new dividend by 10:
320: 5 = 640: 10 = 64.
The proof of this property of the quotient was not carried out in the textbook. In the lesson, it is enough to give him such an example: “Let's prove that if 320: 5 = c, then (320 ∙ 2): (5 ∙ 2) = c, where c is a natural number”.
To do this, we multiply c by 5 ∙ 2 and check if the result is 320 ∙ 2. In this case, we take into account that since 320: 5 = c, then the equality c ∙ 5 = 320 is true.
c ∙ (5 ∙ 2) = (c ∙ 5) ∙ 2 = 320 ∙ 2.
Thus, the property of the quotient is proved for the quotient 320: 5 and natural number c.
Note that if instead of 320 and 5 we take any natural numbers a and b such that the equality a: b = c is true, and instead of 2 we take any natural number d, then, arguing in a similar way, we will obtain a proof of the same statement in general form:
a: b = (a ∙ c): (b ∙ c).
At this point, the tasks are selected so that when solving them, division into a column is not required, which will be studied in paragraph 1.15.
RT. It is advisable to use tasks 91–93 at the initial stage of the study of division. They test the understanding of the division rule (definition). Tasks 94–97 for calculations without a column. Task98 to find unknown components during multiplication and division. Tasks 99–107 to test the understanding of the relationship of components during multiplication and division.
Decisions and comments
188. Prove that if each of the natural numbers a and b is divisible by a natural number c, then the equality (a + b): c = a: c + b: c is true.
Solution. Let us give a general proof. Since each of the natural numbers a and b is divisible by a natural number c, there are natural numbers a: c and b: c. We multiply their sum by c and transform the resulting product using the distribution law and the definition of the quotient (a: c is the number that, when multiplied by c, gives a, therefore (a: c) ∙ c = a):
(a: c + b: c) ∙ c = (a: c) ∙ c + (b: c) ∙ c = a + b,
therefore, the equality (a + b): c = a: c + b: c is true.
If the teacher believes that in his class the general proof (in letters) given by the students is not yet ready to perceive, then it is better to cite it for a specific case, for example, this: (15+ 35): 5 = 15: 5 + 35: 5. However one should not carry out the proof by means of calculations - make sure that the left and right will get the same answer (such a "proof" will not work in letters). It is necessary, albeit on specific numbers, to carry out the same reasoning as in the proof in the general case, this will gradually teach students to prove the statements.
Intermediate control. DM. C – 3.
1.13. Solving word problems using multiplication and division
At this point, the work begun earlier on teaching schoolchildren to solve problems by arithmetic methods continues. In the educational text, the tasks are solved with explanations, but from time to time it is necessary to give the students an instruction: "And this task must be solved with questions." Particular attention should be paid to the fact that some students from elementary school have entrenched misconceptions about the choice of action to solve the problem. If they encounter the question “how much?” In the text of the problem, then they say that it is necessary to subtract, etc. Therefore, task193 must be completed in the classroom and make sure that the actions for getting the answer are chosen correctly.
RT. Problems 108-117 can be used in the first lessons on the topic, solving problems 108-112 with questions, and problems 113-117 with explanations. The solution of problems 118–137 involves the use of all the studied actions.
Decisions and comments
193. a) Each cart was loaded with 8 bags of potatoes. How many carts did the 72 sacks load?
b) Some of the 40 bags were filled with granulated sugar. There are 10 empty packages left. How many bags were granulated sugar poured into?
c) In the sewing workshop there are 2 pieces of cloth, 60 m each. How many meters of cloth are left?
SOLUTIONS TO PROBLEMS
General notes on verification.
The criteria are written on the basis of the solution "reduced" to the problem.
In the case of a “different” solution, other criteria must be developed in accordance with the general requirements for the criteria.
1. Tanya went to buy pens and pencils. Having spent all the money, she could buy 6 pens or 12 pencils. She decided to buy both equally with all the money. How many?
Answer: 4.
Solution.
One pen is like two pencils, and a pen and pencil are like three pencils. Therefore, Tanya can buy 12: 3 = 4 sets of pen and pencil.
Verification criteria.
Based on a specific numerical example: 1 point
2. Twins Anya, Manya and Tanya baked cakes for their birthday. If Anya and Manya had baked twice as many cakes, the total number of cakes would have increased by 60%. What percentage of cakes did Tanya bake?
Solution. If Tanya also baked twice as many cakes, then all the cakes would have increased by 100%. The share of Ani and Mani is 60%, which means that the share of Tanya is 100% -60% = 40%.
Verification criteria.
No reasonable progress, but there is an answer: 0 points
A special case is considered: 1 point.
There is an action of 100% -60%, but an assumption about Tanya is not made: 2 points
3. Four natural numbers were written on the board. By adding them in all possible different ways in two, Petya got the following six sums: 17, 18, 20, 21, 23, 26. Prove that Petya made a mistake when calculating the sums.
Solution. The sum of all six pairwise sums is 125. Each of the numbers written on the board is included in this sum three times, which means that this sum must be a multiple of 3, but 125 is not divisible by 3.
Verification criteria:
Found the sum of all pairwise sums equal to 125: 1 point.
It is indicated that each number is used as a term three times: 2 points.
Both previous statements are made: 3 points
It is noticed that since each number is a summand three times, then the sum should be divisible by 3, but the conclusion that they came to a contradiction is not made: 6 points.
The presence of all the details in the solution: 7 points.
Method 2. Let's arrange the written numbers in non-decreasing order: a £ b £ c £ d. Then
a + b = 17, a + c = 18, b + d = 23, c + d = 26. 18 + 23 = a + b + c + d = 17 + 26. (or 26–23 = c – b = 18–17) We got a contradiction, therefore, there was an error in the calculations.
This solution is presented to demonstrate the fact that the condition "natural numbers" is superfluous. It is for teaching children a different approach to the task (the method of the extreme).
4. Petya has a 5 × 7 rectangle and a 1 × 1 square. Can Petya cut this rectangle into 2 parts that are not rectangles, and then add a 6 × 6 square from these two parts and this 1 × 1 square? (If possible, it should be shown how the rectangle is cut and how the square is made. Or it should be explained why this is not possible.)
Answer. Maybe.
Multiple rectangle cuts and square assemblies are specified.
(There are also other solutions.)
Picture 1
Figure 2.
Figure 3
Figure 4.
Verification criteria:
If there is cutting, but there is only one drawing, that is, it is shown how to assemble or how to cut: 4 points.
5. Six friends: Andrey, Vitya, Borya, Sasha, Tolya and Gena, lined up in a row in descending order of their height (none of them have the same height). Then Gena and Andrey changed places, Borya and Vitya also changed places and, finally, Sasha and Tolya also changed places. It turned out that now the boys are in ascending order of their height. Find the tallest among the boys if it is known that Borya is higher than Andrei and Gena, but lower than Sasha.
Solution... Since after all the permutations, the guys lined up in the opposite order, the tallest and smallest were reversed (1). This pair cannot include Andrey and Gena: they are both lower than Bori (2). Borya cannot enter this pair. He is lower than Sasha, but higher than Andrey, which means he is not the tallest and not the lowest (3). There is only one pair left: Sasha and Tolya. Sasha is taller than Bory and cannot be the lowest (4). This means that the highest is Sasha, and the lowest is Tolya.
Verification criteria:
Only the correct answer is indicated: 1 point.
There is the first statement (1): 2 points.
There are statements (1) and (2): 3 points.
There are statements (1) and (2) and (3): 6 points.
All statements are there: 7 points.
6. There are 10 checkers on a 1´20 strip on 10 left fields. A checker can move to a free cell adjacent to the right or jump over a checker adjacent to the right to the next cell after it, if this cell is free. Left movement is not permitted. Is it possible to rearrange all the checkers in a row without spaces in reverse order?
Solution... Let's number the checkers with numbers 1,2,3, ..., 9,10.
Example permutations. Movements consist of two parts: moving odd (disassembling) and moving even (assembling).
Verification criteria:
All permutations are indicated: 7 points.
Start and end are specified, but there are ellipsis. 6 points.
If it is also there, but there is a skip of moves - 5 points.
Comment. The movement of each piece is shown by the start and end position, intermediate moves are easy to recover. There is no need to find fault with such passes.
1. In the number written on the board, Petya erased three numbers and got a multiple of 9. What number is now written on the board? (List all possibilities and prove that there are no others.)
A number is divisible by 9 only if the sum of its digits is divisible by 9. The sum of the digits of the written number is 30. The sum of three digits from 1 to 3 can vary from 3 to 9. Therefore, after striking out three digits, the sum of the digits of the new number can be from 23 to 27. Of these, only 27 are divisible by 9. This means that three digits have been crossed out, the sum of which is 3, that is, three units. The number will remain on the board:.
Verification criteria.
Answer presented: 1 point.
It is indicated that the divisibility of the sum of the digits by 9 is needed, so you need to cross out three digits, the sum of which is 3, which means that these are three units: 4 points.
For a complete solution, it must be shown that no other sum of numbers, divisible by 9, can be obtained. If this is done - 7 points. If the reasoning shows that three ones are crossed out, but the number is not shown: minus 1 point.
2. Natasha and Inna each bought the same box of tea bags. They are known to brew two or three cups of tea with a single bag. This box was enough for Natasha for 53 cups of tea, and for Inna for 76. How many bags were in the box? The answer must be substantiated.
Solution
Note that the box could not contain less than 26 sachets: if there are at least 25 of them, then Inna will not be able to drink more = 75 cups, but she drank 76.On the other hand, in the box
there could not be more than 26 sachets: if there are at least 27 of them, then Natasha could not drink less = 54 cups, but she drank 53. Thus, there were 26 sachets in the box: Inna brewed 24 sachets three times and 2 sachets twice , and Natasha brewed 1 sachet three times and 25 sachets twice.
Verification criteria.
Provided only answer 26 sachets: 0 points.
It is imperative that you show a way to drink 53 and 76 cups of tea, otherwise the solution will not be complete. Missing each example: minus 1 point.
3. Seven dwarfs of various ages are sitting behind round table... It is known that every dwarf can speak truth or lie. Each of them said that he is older than his neighbors. What is the largest number of true statements that could be?
Grade. Consider a senior gnome. He couldn't tell the truth. Divide the remaining 6 into three adjacent pairs. In each pair, only one gnome could tell the truth. It means that no more than three dwarfs told the truth. Example: 7, 5, 6, 3, 4, 1, 2. (Dwarves are numbered by seniority.)
Verification criteria.
Assessment problem plus an example.
Example: 2 points.
Rating: 4 points.
When assessing, it is important that neighboring gnomes cannot both speak the truth, and if at least four speak the truth, then there are neighbors among them.
All together 7 points.
Comment. If the gnomes sat in a row, then 4 gnomes could tell the truth.
6, 7, 4, 5, 2, 3, 1.
4. It is known that . Find .
Solution
Let's add the fractions on the left side:
Where does it mean 
... Once again, adding the fractions on the left side of the last equality, we get.
Finally, we have 
5. Little children ate sweets. Each ate 11 fewer candies than everyone else combined, but still more than one candy. How many candies were eaten in total?
Solution
Let's choose one of the children - for example, Petya. If you take 11 of all the other candies, there will be the same amount as Petya's. This means that twice the number of Petya's candies is equal to the total number of candies minus eleven. The same can be said about any of the children, which means that all children have an equal share of sweets - say, one heap.
It is clear that everyone ate an integer number of heaps less than the others together. Therefore, 11 is divided by the pile size. This means (since, according to the condition, everyone ate more than 1 candy), there are 11 candies in piles, that is, everyone ate a pile less than the others together. Petya ate one pile, therefore, the rest - two. This means that there are three heaps in total, and 33 chocolates.
The same solution can also be written algebraically.
Let us denote by S the total number of sweets the children ate. If one of the children ate a sweets, then by condition everyone else ate a + 11 candies, and thus all ate together S = a +(a + 11)=
2a + 11 sweets. This reasoning is true for every child, so all children ate the same amount of candy: a =(S– 11)/
2 pieces.
We now denote by N number of children. Then the condition is written as a = a(N– 1)–
11, whence 11 = a(N– 2). The number 11 is simple, so one of the factors is 1, and the other 11. But by condition a> 1, therefore a = 11 , N– 2=
1 . Thereby N = 3, and was eaten S = aN = 33 candies.
Answer: 33 candies.
Only answer: 0 points.
6. Points K and D were taken on the sides AB and AC of triangle ABC, respectively. Point E was chosen so that K is the midpoint of the segment DE. It turned out that РEAK = РACB and AE = DC. Prove that BD is the bisector of angle ABC.
From point D we drop perpendiculars DL and DM to lines AB and BC, respectively. From point E we drop the perpendicular EN to line AB. Right-angled triangles AEN and CDM are equal in hypotenuse and acute angle. Hence DM = EN. In addition, EN = DL (from the equality of right-angled triangles, if N and L are different, or as coinciding with the segments EK and DK, if the points N, L and K coincide).
Hence DL = DM, and point D is equidistant from the sides of angle ABC and, therefore, lies on the bisector of this angle.
Verification criteria. Desired perpendiculars omitted: 1 point.
When proving the equality EN = DL, the case of coincidence of the bases of perpendiculars was not considered: minus 1 point.
1. Natural number cube N is divisible by 2010. Does it follow that the number itself N divisible by 2010? Answer: it should.
Solution... 2010 = 2 * 3 * 5 * 67. The numbers 2, 3, 5 and 67 are prime.
https://pandia.ru/text/77/496/images/image018_66.gif "width =" 19 height = 15 "height =" 15 ">. gif" width = "21" height = "21 src ="> divisible by 3 divisible by 3,
https://pandia.ru/text/77/496/images/image018_66.gif "width =" 19 "height =" 15 "> divisible by 5,
https://pandia.ru/text/77/496/images/image018_66.gif "width =" 19 "height =" 15 "> is divisible by 67.
Only answer given: 0 points.
2. There are cans of different sizes: A, B, C and D. It is known that 11 cans A and 7 cans B hold the same amount as 12 cans C. C 6 cans A and 5 cans B hold the same amount as 6 cans C and 1 can D. 6 cans D are completely filled with water. Will 3 cans A and 8 cans B be enough to pour all the water from 6 cans D?
Solution. Let https://pandia.ru/text/77/496/images/image021_51.gif "height =" 15 src = "> be the volumes of cans A, B, C and D, respectively.
For a correctly composed system of equations: 2 points.
3.
Given a parallelogram KLMN acute apex K... On the rays KL and ML marked points A and B respectively, and AM = LM and BK = KL.
a) Prove that AN = BN.
b) Prove that triangles ABN and BKL are similar.
Solution.
From the equality of triangles AMN and BKN(on two sides and the angle between them) follows the equality of the segments AN and BN.
From equality of angles AKB and AMB(the angles at the vertices of similar isosceles triangles BKL and AML) it follows that the points A, B, K, M lie on the same circle, and since
then the point also lies on this circle N... Therefore, the angles BNA and BKL at the tops N and K isosceles triangles BNA and BKL are equal. Therefore, triangles are similar.
Point a) is proved: 3 points.
Point b) is proved: 4 points.
4. Prove that if the equations and https://pandia.ru/text/77/496/images/image028_31.gif "width =" 263 "height =" 24 "> have no roots.
Solution.
Let's take an arbitrary one.
Then it has no roots, so for any.
The equation has no roots, therefore, for any. Therefore, for anyone.
https://pandia.ru/text/77/496/images/image034_29.gif "width =" 255 "height =" 22 src = ">
for anyone . That is, the equation
It is proved that for any +4 points.
If there is no corresponding explanation, then there is no corresponding addition of points.
5. Vasya forgot the four-digit code in the storage room (the code can be anything from 0000 to 9999). He only remembers that the number that specifies the code is divisible by 3 and 7 and not divisible by 5 and 9. How many options will he have to go over to be sure to guess the code?
Answer: 254.
Solution. 1 way.
Code 0000 does not work.
Among the numbers from 1 to 9999, exactly = 476 are divisible by 21..gif "width =" 65 "height =" 39 ">. But among 158 numbers divisible by 9 and among 95 numbers divisible by 5, there are coincidences. numbers divisible by 45. There are exactly such numbers out of 476 numbers divisible by 21. Then there are exactly + 31 = 254 numbers satisfying the condition of the problem.
9 * 5 * 7 = 315, therefore, among the numbers from 1 to 315, from 316 to 630, from 630 to 945, etc., there is the same number of numbers that satisfy the condition of the problem. There are exactly 8 such numbers from 1 to 315 (these are the numbers 21, 42, 84, 147, 168, 231, 273, 294). Hence, from 1 to 315 * 31 = 9765 such numbers 31 * 8 = 248. It remains to consider the numbers from 9766 to 9999 and make sure that among them exactly 6 numbers satisfy the condition of the problem (9786, 9807, 9849, 9912, 9933, 9996). Total 248 + 6 = 254 numbers.
Answer without solution: 0 points.
The indicated formula is + 31 = 254: + 3 points.
Each computational error: - 1 point.
Answer without solution: 0 points.
It is indicated that among each of the following 315 numbers the same number of numbers satisfying the condition of the problem: +3 points.
It is calculated that exactly 8 from 1 to 315 satisfies the condition: +1 point.
It is calculated that exactly 6 from 9766 to 9999 satisfy the condition of the problem: +1 point.
The formula is like 248 + 6 = 254: +2 points.
If someone has the patience to write down all 254 numbers and not be mistaken: 7 points.
6.
Points A and B are taken on the graph of the function 
... Of these, the perpendiculars are omitted on the abscissa axis, the bases of the perpendiculars are HA and HB; С is the origin. Prove that the area of the figure bounded by straight lines CA, CB and arc AB is equal to the area of the figure bounded by straight lines AHA, BHB, abscissa axis and arc AB. 5.
Points A and B are taken on the graph of the function. Of these, the perpendiculars are omitted on the abscissa axis, the bases of the perpendiculars are HA and HB; С is the origin. Prove that the area of the figure bounded by straight lines CA, CB and arc AB is equal to the area of the figure bounded by straight lines AHA, BHB, abscissa axis and arc AB.
Solution. We can assume that the abscissa of point A is less than the abscissa of point B (see Fig.). Consider the point K of intersection of the segments AHA and СB. Then the difference of the considered areas is equal to the difference between the areas of the triangle СAK and the quadrangle HAKBHB, which, in turn, is equal to the difference between the areas of the triangles СAHA and СBHB. And since СHA * AHA = СHB * BHB = 2010 (A and B are on the graph), these areas are equal to each other.
It is shown that the difference between the areas under consideration is https://pandia.ru/text/77/496/images/image044_20.gif "width =" 101 "height =" 23 src = ">: 4 points.
It is proved that either 
: +3 points.
1. ... Prove that for all natural numbers the inequality
Solution: Divide both sides of the inequality by a positive value. Get the inequality If then the degree is negative and the inequality is true .. gif "width =" 61 "height =" 19 ">: 0 points.
Received view 
or: 1 point
2. Can the sum of the digits for some natural k be the same for the next two numbers https://pandia.ru/text/77/496/images/image059_10.gif "width =" 76 "height =" 24 src = ">?
Answer: it cannot.
Solution. Let's denote https://pandia.ru/text/77/496/images/image061_11.gif "width =" 171 "height =" 24 src = ">. One of three consecutive numbers is divisible by three, therefore, one of the numbers is https : //pandia.ru/text/77/496/images/image064_9.gif "width =" 53 "height =" 21 src = "> divisible by three, and the other is not. Therefore, the sum of the digits of only one of them is divisible by three. Hence, they are different.
3. Square trinomials and positive real roots x 1, x 2 and x 3, x 4, respectively, and x1 < x3 < x2 < x4 . Prove that a square trinomial https://pandia.ru/text/77/496/images/image068_9.gif "width =" 85 "height =" 51 src = ">. Gif" width = "111" height = "21 ">.
A different justification for the inequalities is possible - a<-c, b<d using the properties of a quadratic function.
A solution is given, but when passing from the inequalities - a<-c and b<d to inequalities a 2<c 2, 4b2 <4d2 not substantiated that - a, b,-c, d positive: 5 points.
4. 2010 zeros are inserted between every two digits of 1331. Prove that the resulting number is divisible by 1331.
Solution. Let's imagine the number https://pandia.ru/text/77/496/images/image072_8.gif "width =" 386 "height =" 24 ">
https://pandia.ru/text/77/496/images/image074_8.gif "width =" 55 "height =" 35 src = "> divisible by 11 (divisible by 11), which means 100..0013 divisible by 113 = 1331.
The number is presented in the form https://pandia.ru/text/77/496/images/image076_8.gif "width =" 31 "height =" 24 ">.
Solution.
Let be O Is the center of the circle, since D AB C is isosceles, then BO=OC... Consider D FBO and D ECO: Ð FBO=Ð ECO= a, Bf× CE=6, BO× OC=BC 2/4 = 6, that is Bf×
CE=BO×
OCÛ https://pandia.ru/text/77/496/images/image079_8.gif "width =" 103 height = 38 "height =" 38 ">. Since Ð BOF= b, l EOC= g, then РFOE = a. From the equalities BO=
OC and 
follows that . Consider D FOE and D ECO:
Ð FOE=Ð ECO= a, and https://pandia.ru/text/77/496/images/image047_16.gif "width =" 13 height = 15 "height =" 15 "> inequality is true
Solution: Divide both sides of the inequality by a positive value. Get the inequality If then the degree is negative and the inequality is true .. gif "width =" 61 "height =" 19 ">: 0 points.
Received view or: 1 point
Proven for only one of the cases or: 3 points.
2. Solve the equation: https://pandia.ru/text/77/496/images/image082_8.gif "height =" 20 src = ">
Answer: There are no solutions.
First solution: The sequence https://pandia.ru/text/77/496/images/image084_7.gif "width =" 31 "height =" 21 "> .. gif" width = "13" height = "15"> is not equal to 0. The equation has no roots.
If it is noticed that this is the sum of a geometric progression: 1 point
Amount found, but no conclusion was made: +1 point.
Replacement made: 1 point.
Second solution: Not a solution to the equation. Divide both sides of the equation by and get the equation.
We rewrite the terms in the following order
https://pandia.ru/text/77/496/images/image092_6.gif "width =" 75 "height =" 21 "> .. gif" width = "84" height = "24"> which has roots 
, 
Note that according to the Cauchy inequality 
and therefore both roots do not fit.
