The sum of 100 different natural numbers is 5130
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 (perhaps one), the numbers on the board were written less than the original by one. The numbers, which then turned out to be 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 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.
Decision.
and) 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 became 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.
at) 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
There are 100 different natural numbers written on the board with the sum of 5120.
a) Can the number 230 be written?
b) Is it possible to do without the number 14?
c) What is the smallest number of multiples of 14 on the board?
Decision.
a) Let the number 230 and 99 other different natural numbers be written on the board. The minimum possible sum of the numbers on the board is achieved, provided that the sum of 99 different natural numbers is minimal. And this, in turn, is possible if 99 different natural numbers are an arithmetic progression with the first term and the difference.The sum of these numbers, according to the formula for the sum of an arithmetic progression, is:
The sum of all numbers on the board S will be equal to:
It is easy to see that the resulting sum is greater than 5120, which means that any sum of 100 different natural numbers, among which there are 230, is greater than 5120, therefore, 230 cannot be on the board.
b) Suppose the number 14 is not written on the board. In this case, the minimum possible amount S numbers on the board will consist of two sums of arithmetic progressions: the sum of the first 13 members of the progression with the first member, the difference (that is, the series 1,2,3, .. 13) and the sum of the first 87 members of the progression with the first member, the difference (that is, the series 15,16,17, .. 101). Let's find this amount:
It is easy to see that the resulting sum is greater than 5120, which means that any sum of 100 different natural numbers, among which there are no 14, is greater than 5120, therefore, one cannot do without the number 14 on the board.
c) Suppose that all the numbers from 1 to 100 are written on the board. Then it turns out that the resulting series is an arithmetic progression with the first term, the difference. By the formula for the sum of an arithmetic progression, we find the sum of all numbers on the board:
The received amount does not satisfy the condition of the problem. Now, in order to increase the sum of all the numbers written on the board to the one indicated in the condition, we will try to replace the numbers that are multiples of 14 with other numbers following a hundred: 70 will be replaced by 110, 84 - by 104, and 98 - by 108. The resulting sum S will be equal to:
With the further replacement of numbers that are multiples of 14 with numbers greater than 100, the amount will increase and will not correspond to the condition of the problem. So the smallest number of multiples of 14 is 4.
Let us give another solution to item c).
Let's give an example, when on the blackboard there are four numbers that are multiples of 14 (14, 28, 42, 56):
1, 2, ... , 69, 71, 72, ... , 83, 85, 86, ... , 97, 100, 101, 102, 103, 115.
Let us prove that there cannot be three numbers that are multiples of 14. To remove the maximum number of numbers that are multiples of 14, it is necessary that the differences between the new and old numbers be minimal. That is, it is necessary to replace the largest numbers, multiples of 14, with the smallest possible, greater than one hundred numbers. Let the number of multiples of 14 be 3. Then the minimum sum of the numbers written on the board is:
The resulting sum is greater than 5120. With further replacement of numbers that are multiples of 14 with numbers greater than 100, the amount will increase, which means that there cannot be less than four numbers on the board that are multiples of 14.
A) No b) No c) 4.
Posted on 03/14/2018
5 (100%) 1 vote
There are 100 different natural numbers written on the board, and it is known that the sum of these numbers is 5120.
a) Could 230 be written on the board?
b) Could it be that the number 14 is not written on the board?
c) What is the smallest number of multiples of 14 on the board?
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Sadne-ss
2 minutes ago
and) Let's calculate the option in which the amount will be the smallest. Naturally, this is just the sum of the first hundred numbers, i.e. 1+2+3…+100
... You can count by sorting out, or you can use the formula " arithmetic progression sums".
Now we calculate the amount. S100 \u003d ((1 + 100) / 2) * 1-00 \u003d 5050;
We need to try somehow, replace any number in our row with 230 ... We find out how much we are missing before the given in the condition: 5120-5050=70 , yeah, and what was the largest number in our row? Correctly, 100 ... It turns out that the largest number with which we can replace any number from our series is 170 ... So the numbers 230 in a row it can't be.
No answer;
b) Take, all the same row, from 1 to 100, but remove the number from there 14 and try to replace it with another one. For example, let's try to take the smallest number after 100 , namely 101 and we will replace it. The sum of the first hundred numbers we found, which means that for replacement, we need subtract from it 14 and add new value 101: 5050-14+101=5137 -. Unfortunately, the condition says that the sum is 5120 , so alas, you cannot exclude the number 14 from our list.
Answer: b) No;
at) Find all multiples 14 from our series ( from 1 to 100). There are many ways to find multiple values, but in our case, the number is not so large, they can be sorted out manually, we get a series by adding: 14, 28, 42, 56, 70, 84, 98 . Only 7 multiples of 14... Now let's try to replace them with more large values \u200b\u200bare not multiples of 14because at the moment, our sum is 5050... Replace the largest multiple with the smallest unused: 98 to 101;
Our amount will become: (101-98)+5050=5053- ;
Amount: (102-84) + 5053 \u003d 5071-;
There is still a place, we continue. Replace 70 with 103;
Amount: (103-70) + 5071 \u003d 5104-;
5104 , still less than 5120, then let's move on. Replace 56 with 104;
Amount: (104-56) + 5104 \u003d 5152-;
It turned out more than necessary, which means you need
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