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I've found already that it's only true when the total amount of numbers is an odd number. I've also found that the median and mean are the same with consecutive numbers. I can not prove that the it's only true with odd numbers.

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In summary: Half of an odd number is not an integer. So n must be odd, and the expression has a remainder of n/2.In summary, the statement "The sum of n consecutive numbers is always divisible by n" is only true when the total amount of numbers is an odd number. This is because when n is even, the expression has a remainder of n/2. Additionally, it has been found that the median and mean are the same with consecutive numbers, but it has not been proven that this is only true with odd numbers. To find the sum of consecutive numbers, a general formula can be used by summing the remainders after dividing by n.

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I've found already that it's only true when the total amount of numbers is an odd number. I've also found that the median and mean are the same with consecutive numbers. I can not prove that the it's only true with odd numbers.

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try writing down a general formula for the sumMatt said:

I've found already that it's only true when the total amount of numbers is an odd number. I've also found that the median and mean are the same with consecutive numbers. I can not prove that the it's only true with odd numbers.

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The remainders, after dividing by n, for n consecutive numbers are 0,1,...,n-1. Sum the remainders to get n(n-1)/2.

If n is odd, n-1 is even and (n-1)/2 is an integer, so this sum is divisible by n.

If n is even, n-1 is odd. n(n-1)/2 = n(n-2)/2 + n/2. (n-2)/2 is an integer, so n/2 is the remainder after dividing the sum by n.

If n is odd, n-1 is even and (n-1)/2 is an integer, so this sum is divisible by n.

If n is even, n-1 is odd. n(n-1)/2 = n(n-2)/2 + n/2. (n-2)/2 is an integer, so n/2 is the remainder after dividing the sum by n.

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You missed the point. In your derivation 2k-n-1 must be even, so n must be odd. When n is even the expression has a remainder of n/2.HallsofIvy said:

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Mathman, could I run a couple of problems past you if you have a moment?

Regards,

Matt.

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Thank you very much for your everyone. I really appreciate it.mathman said:You missed the point. In your derivation 2k-n-1 must be even, so n must be odd. When n is even the expression has a remainder of n/2.

Mathman, could I run a couple of problems past you if you have a moment?

Regards,

Matt.

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If you submit them via the forum I will look at them. I prefer not to answer privately.Matt said:

Mathman, could I run a couple of problems past you if you have a moment?

Regards,

Matt.

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How do you get that?mathman said:You missed the point. In your derivation 2k-n-1 must be even,

so n must be odd. When n is even the expression has a remainder of n/2.

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2k--n-1: 2k is even, if n is even, n+1 is odd. Even number minus odd number is odd.

When the sum of n consecutive numbers is divisible by n, it means that the result of the sum divided by n will have no remainder. In other words, the sum can be divided evenly by n.

The formula for finding the sum of n consecutive numbers is (n/2)(2a + (n-1)d), where "a" is the first number in the sequence and "d" is the common difference between each consecutive number.

To prove that the sum of n consecutive numbers is divisible by n, you can use mathematical induction. This involves showing that the formula for finding the sum (as mentioned in the previous question) holds true for any value of n, and then using this to show that it also holds true for n+1.

Yes, the sum of n consecutive numbers can be divisible by n even if n is an odd number. This is because the formula for finding the sum (as mentioned in the second question) takes into account the common difference between each consecutive number, which can be any integer regardless of whether n is odd or even.

Understanding sums of consecutive numbers being divisible by n can be useful in various fields such as cryptography, computer science, and data analysis. For example, in cryptography, this concept is used in creating secure algorithms for encryption and decryption. In computer science, it can be used in optimizing algorithms and data structures. In data analysis, it can be used to identify patterns and trends in numerical data.

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