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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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- Thread starter Matt
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- #1

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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.

- #2

lavinia

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try writing down a general formula for the sum

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.

- #3

mathman

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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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HallsofIvy

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mathman

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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.

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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.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.

- #8

mathman

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

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

Regards,

Matt.

- #9

HallsofIvy

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How do you get that?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.

- #10

mathman

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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.

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