There are 100 different things written on the blackboard. natural numbers 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 that can be on the board?
Solution.
a) Let the number 230 and 99 other different natural numbers be written on the board. The minimum possible sum of numbers on the board is achieved under the condition 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 member and the difference The sum of these numbers, according to the sum formula arithmetic progression, will be:
The sum of all the 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, the number 230 cannot be on the board.
b) Let the number 14 not be 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 terms of the progression with the first term, difference (that is, the series 1,2,3,..13) and the sum of the first 87 terms of the progression with the first member, 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 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 member, the difference. Using the formula for the sum of an arithmetic progression, we find the sum of all numbers on the board:
The resulting 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, let's try to replace the numbers that are multiples of 14 with other numbers following the hundred: 70 will be replaced by 110, 84 by 104, and 98 by 108. The resulting sum S will be equal to:
With further replacement of numbers that are multiples of 14 by numbers greater than 100, the sum will increase and will not correspond to the condition of the problem. So the least number of multiples of 14 is 4.
Let us give another solution to part c).
Let's give an example when four numbers that are multiples of 14 (14, 28, 42, 56) are written on the board:
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 maximum amount multiples of 14, it is necessary that the difference between the new and old numbers be minimal. That is, you need to replace largest numbers, multiples of 14, by the smallest possible, greater than one hundred numbers. Let the number of numbers that are 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 by numbers greater than 100, the sum will increase, which means that there cannot be less than four numbers that are multiples of 14 on the board.
A) no b) no c) 4.
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There are 100 different natural numbers written on the blackboard with the sum 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 that can be on the board?
Solution.
a) Let the number 230 and 99 other different natural numbers be written on the board. The minimum possible sum of numbers on the board is achieved under the condition 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 member and a difference. The sum of these numbers, according to the formula for the sum of an arithmetic progression, will be:
The sum of all the 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, the number 230 cannot be on the board.
b) Let the number 14 not be 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 terms of the progression with the first term, difference (that is, the series 1,2,3,..13) and the sum of the first 87 terms of the progression with the first member, 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 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 member, the difference. Using the formula for the sum of an arithmetic progression, we find the sum of all numbers on the board:
The resulting 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, let's try to replace the numbers that are multiples of 14 with other numbers following the hundred: 70 will be replaced by 110, 84 by 104, and 98 by 108. The resulting sum S will be equal to:
With further replacement of numbers that are multiples of 14 by numbers greater than 100, the sum will increase and will not correspond to the condition of the problem. So the least number of multiples of 14 is 4.
Let us give another solution to part c).
Let's give an example when four numbers that are multiples of 14 (14, 28, 42, 56) are written on the board:
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 new and old numbers be minimal. That is, it is necessary to replace the largest numbers, multiples of 14, with the smallest possible numbers, greater than one hundred. Let the number of numbers that are 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 by numbers greater than 100, the sum will increase, which means that there cannot be less than four numbers that are multiples of 14 on the board.
A) no b) no c) 4.
