Article: 946 of rec.puzzles
From: J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger)
Subject: Re: simple number puzzle
Date: 22 Dec 92 01:06:21 GMT
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Yuzuru Hiraga writes:
A simple number puzzle for Christmas...
What positive integer cannot be expressed as a sum of 2 or more
consecutive integers?
I believe the answer is 2^n power (where n is an integer) can't be
expressed as the sum of consecutive integers, but all other numbers can.
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**End of quoted material.
Proof:
1. All odd integers can be expressed as the sum of two consecutive
integers.
2. 4 consecutive integers will give all numbers with exactly one factor
of 2.
3. x number of consecutive integers, where x is prime (and not equal to
2), will give you all multiples of x (starting at x^2+x/2).
Numbers lower than (x^2+x)/2 fall into two cases:
a) Odd numbers, covered above.
b) Even numbers. As these have only one multiple of 2, they will
obviously be a number that has a remainder of 2 when divided by 4 (ie.
case #2 above).
This solves all numbers except for those whose only factors are 2 (ie.
powers of 2).
As I am leaving to go home for Christmas tomorrow morning, anyone who
has the rest of the proof should email me directly with it.
Thanks!
--Ted Schuerzinger
email: .zed@Dartmouth.EDU
"I should have known it would be bad vodka when all the label said was
'Russian Vodka'."
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Article: 957 of rec.puzzles
From: costley@solo.eng.hou.compaq.com (Brett Costley)
Subject: Re: simple number puzzle
Date: Tue, 22 Dec 1992 17:23:38 GMT
In article (J. Theodore Schuerzinger) writes:
>Yuzuru Hiraga writes:
>
>A simple number puzzle for Christmas...
>
>What positive integer cannot be expressed as a sum of 2 or more
>consecutive integers?
>
>I believe the answer is 2^n power (where n is an integer) can't be
>expressed as the sum of consecutive integers, but all other numbers
>can.
>
"proof" deleted
2^0 = 1 = 0+1
2^1 = 2 = -1+0+1+2
2^2 = 4 = (-3)+(-2)+(-1)+0+1+2+3+4
2^n = [-(n-1)]+...+0+...+n
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Article: 949 of rec.puzzles
From: zhang@climat.geology.yale.edu
Subject: Re: simple number puzzle
Date: 22 Dec 92 04:31:57 EDT
In article (Yuzuru Hiraga) writes:
> A simple number puzzle for Christmas...
>
> What positive integer cannot be expressed as a sum of 2 or more
> consecutive integers?
>
> More specifically, given a positive integer, how many ways can it
> be expressed as a sum of 2 or more consecutive [positive] integers
> (including none)?
Clearly, any positive integer number that could be expressed as a sum of
2 or more consecutive integers can be written as either (n+k)(2k+1) or
k(2n+2k-1) where both n and k are any integers that are greater than or
equal to 1. In other words, any integer number which has at least one odd
factor greater than or equal to 3 can be expressed as a sum of 2 or more
consecutive integers. (All the prime numbers greater than 2 are included here.)
The exceptions, therefore, are only powers of 2.
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Article: 954 of rec.puzzles
From: radcliff@csd4.csd.uwm.edu (David G Radcliffe)
Subject: Re: simple number puzzle
Date: 22 Dec 1992 16:48:27 GMT
Given a positive integer n, in how many ways can it be expressed
as the sum of (one or more) consecutive positive integers? The
surprising answer is that the number of such representations is
equal to the number of odd divisors of n.
First we find the number of ways of representing n as the sum of
consecutive integers. To write n as the sum of an odd number of
consecutive integers, we need to solve (a-b) + . . . + (a+b) = n.
That is, a(2b+1)=n. The number of solutions of this equation is
equal to the number of odd divisors of n. Similarly, to write n
as the sum of an even number of consecutive integers, we need to
solve (a-b+1) + . . . + (a+b) = n; that is, (2a+1)b=n. Again,
the number of solutions of this equation is equal to the number of
odd divisors of n. The total number of representations is thus
twice the number of odd divisors of n.
If n=a+...+b then n=(1-a)+...+b. Define an equivalence relation
on the set of all solutions by declaring that {a,...,b} ~ {1-a,...,b}.
Each equivalence class has two elements, and exactly one of these
elements contains only positive integers. Therefore, half of
the solutions to n=a+...+b are positive. So, the number of solutions
to n=a+...+b in positive integers is equal to the number of odd
divisors of n.
--
David Radcliffe radcliff@csd4.csd.uwm.edu