Eleven numbers with the sum 105 are written on the circle. The sum of any two consecutive numbers is not smaller than 18. Find the least possible value of N, so that each number on the circle is not greater than N, for any set of numbers.
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15
To find the least possible value of N, we need to find the maximum possible value for the smallest number on the circle.
Let's assume the smallest number on the circle is x. Since the sum of any two consecutive numbers is not smaller than 18, the largest number on the circle would be (105 - x).
Now, let's consider three consecutive numbers on the circle. The sum of these three numbers would be:
x + (x+1) + (x+2) = 3x + 3
Since the sum of any two consecutive numbers is not smaller than 18, this implies:
3x + 3 ≥ 18
3x ≥ 15
x ≥ 5
So the smallest number on the circle must be at least 5.
Now, let's consider the largest number on the circle:
(105 - x) ≥ 18
x ≤ 87
Therefore, the smallest number on the circle must be at least 5 and the largest number must be at most 87.
To find the least possible value of N, we need to find the maximum value between 5 and 87.
Hence, the least possible value of N is 87.
To find the least possible value of N, we need to determine the maximum value among the eleven numbers written on the circle. Let's break down the problem step by step.
Step 1:
We are given that the sum of any two consecutive numbers is not smaller than 18. Let's assume the eleven numbers written on the circle are a₁, a₂, a₃, ..., a₁₁.
Step 2:
Since the sum of any two consecutive numbers is not smaller than 18, the sum of all eleven numbers must be greater than or equal to 11 * 18 = 198. Therefore, we have the equation: a₁ + a₂ + a₃ + ... + a₁₁ ≥ 198.
Step 3:
We also know that the sum of the eleven numbers is 105. So we have the equation: a₁ + a₂ + a₃ + ... + a₁₁ = 105.
Step 4:
Combining the two equations from Step 2 and Step 3, we get: a₁ + a₂ + a₃ + ... + a₁₁ ≥ 198 and a₁ + a₂ + a₃ + ... + a₁₁ = 105.
Step 5:
To find the least possible value of N, we need to determine the maximum value among the eleven numbers. In other words, N is the largest number among a₁, a₂, a₃, ..., a₁₁.
Step 6:
Given that the sum of all eleven numbers is 105, the average value of the eleven numbers is 105/11 = 9.545 (rounded to three decimal places).
Step 7:
In order for the average value to be maximized, we want the largest possible numbers to be towards the ends of the circle. Since the numbers are written on the circle, we can imagine the largest number being followed by the smallest number, and so on.
Step 8:
To maximize the average, let's assign the largest number to a₁ and the smallest number to a₂. This means a₁ + a₂ = 18 (because the sum of any two consecutive numbers is not smaller than 18).
Step 9:
Substituting a₁ = 18 - a₂ in the equation a₁ + a₂ + a₃ + ... + a₁₁ = 105, we get (18 - a₂) + a₂ + a₃ + ... + a₁₁ = 105.
Step 10:
Rearranging the terms, we have: 18 + (a₃ + a₄ + ... + a₁₁) = 105.
Step 11:
Simplifying, we get: a₃ + a₄ + ... + a₁₁ = 105 - 18 = 87.
Step 12:
Since we want to maximize the average, we want the remaining numbers, from a₃ to a₁₁, to be as large as possible. Therefore, we assign the value a₃ = a₄ = ... = a₁₁ = N.
Step 13:
From step 11, we know that a₃ + a₄ + ... + a₁₁ = 87. Since a₃, a₄, ..., a₁₁ are all equal to N, we have 9N = 87.
Step 14:
Solving for N, we get N = 87/9 = 9.667 (rounded to three decimal places).
Step 15:
The least possible value of N, such that each number on the circle is not greater than N for any set of numbers, is 9.667 (rounded to three decimal places).
Therefore, the answer is N = 9.667.