Consider numbers from 1 to 2023. We want to delete 3 consecutive, so that the avarage of the left numbers is a whole number. How do we do that



Answer to a math question Consider numbers from 1 to 2023. We want to delete 3 consecutive, so that the avarage of the left numbers is a whole number. How do we do that

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Determine the average of the range 1 to 2023. The sum of consecutive integers from 1 to N is given by the formula: sum = (N/2) x (first number + last number). In this case, N = 2023, and the first number is 1, and the last number is 2023. So the sum is (2023/2) (1 + 2023) = 2,047,276. Since the average of the remaining numbers needs to be a whole number, the sum of the remaining numbers must be divisible by the number of remaining numbers. The average of the three numbers to be removed must be whole numbers: \frac{(x+x+1+x+2)}{3} = 2023 3x + 3 = 3(2023) 3x = 6066 x = 2022 x is the first number the numbers are: 2022, 2023, 1

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