I was working on an extra credit problem dealing with summations for my Algebra 2 class when I came across a conjecture I%26#039;ve yet to prove. For any positive odd number x, and for any n which is a factor of 2x, then x can be expressed as a sum of n consecutive integers.
As an example, suppose x = 5.
n is any factor of 10, meaning it%26#039;s 1, 2, 5, or 10.
Therefore, 5 can be expressed as a sum of 1, 2, 5, or 10 consecutive integers.
5 = 5. [n = 1, 1 addend.]
5 = 2 + 3. [n = 2, 2 addends.]
5 = -1 + 0 + 1 + 2 + 3. [n = 5, 5 addends.]
5 = -4 + -3 + -2 + ... + 5. [n = 10, 10 addends.]
Any real math geeks out there who care to share in proving this conjecture is true? Thanks!
Real math geeks only, please...?
Actually, it%26#039;s fairly easy to show this for quite a larger class of numbers.
Let x be any number, and let n be an odd factor of x. Then x can be written as the sum of n consecutive numbers.
PROOF:
We want to find b (the first number in the sequence) such that
x = 鈭?j=0 to n-1) (b + j).
Let%26#039;s twiddle: then
x = 鈭?j=1 to n) (b-1+j)
= 鈭?j=1 to n) (b-1) + 鈭?j=1 to n) j
= n(b-1) + n(n+1)/2.
Now, solve for b:
nb = x + n - n(n+1)/2
b = x/n + 1 - (n+1)/2.
So long as n divides x and (n+1) is even (i.e. n is odd), this gives an integer solution for b.
搂
** REMARK: **
Actually, the proof shows slightly more... If (x/n - 1/2) is an integer and n is even, this also provides an integer solution. This is the same as saying that n is an even number dividing 2x but not x. For example, if x=5 and n=2 or n=10, this gives b=2 and b=-4 respectively. Compare this to your case above.
Real math geeks only, please...?
Looks interesting, maybe it can be proven by induction, if it is true for 2m+1 where m%26gt;=0 then probably we can prove for (2m+1)+2, what do you think?
Real math geeks only, please...?
x = 鈭?j=0 to n-1) (b + j).
x = 鈭?j=1 to n) (b-1+j)= n(b-1) + n(n+1)/2.
nb = x + n - n(n+1)/2
b = x/n + 1 - (n+1)/2.
.. this is not at all a conjecture..
try the Riemann hypothesis !
Real math geeks only, please...?
factors of 2x = 2x, 1 and other factors a %26amp; a%26#039; etc
x is positive odd number.
factor 1 you just pick the number
factor 2x you pick x-1 consecutive negative numbers x -1 positive numbers, 0 inbetween and x which is consecutive to x-1. and for factor a they need to have an average of a%26#039; and as it is odd pick a%26#039; and a-1/2 before and after it. vice versa for a%26#039;. hope that can help. sorry didnt do it very algebraicly but its difficult on the computer.
Real math geeks only, please...?
If x is prime as in your example, it should be easy to prove since the only factors are 1, 2, x, and 2x.
1 --- x = x
2 --- (x-1)/2 + (x+1)/2 = x
x --- let k =(x+1)/2
k + k-1 + k-2 + ... + k-(x-1)
= xk - x(x-1)/2
= x(x+1)/2 - x(x-1)/2
= x
2x --- 1-x + 1-x+1 + ... + x-2 + x-1 + x
= -(x-1) + -(x-2) + ... + 0 + ... + (x-2) + (x-1) + x
= x
For a composite odd number, let y | x.
Since x is odd, y must be odd. Then the sum of integers centered around (x/y) adds up to x:
[(x/y)-(y-1)/2] + [(x/y)-(y-1)/2+1] + ... + (x/y) + ... + [(x/y)+(y-1)/2 - 1] + [(x/y)-(y-1)/2]
= y(x/y)
= x
Also, we need to prove for 2y. Suppose there exists a series. Let%26#039;s try to find it.
Let k + (k+1) + ... + (k+2y-1) = x.
2yk + 2y(2y-1)/2 = x
2k + (2y-1) = x/y
k = (x/y+1)/2 - y
Since x and y are odd, (x/y+1)/2 is an integer, so k is an integer. Therefore an integer solution exists.
I hope this is correct. Better double check it.
Real math geeks only, please...?
As an example, suppose x = 6.
n is any factor of 12, meaning it%26#039;s 1, 2, 3, 6 or 12.
Therefore, 6 can be expressed as a sum of 1, 2, 3, 6 or 12 consecutive integers.
6 = 6. [n = 1, 1 addend.]
6 = 2 + 4 [n = 2, 2 addend.]
6 = -2 - 1 + 0 + 1 + 2 + 3 [n = 3, 3 addend.]
6 = -5 - 4 - 3 - 2 - 1 + 0 + 1 + 2 + 3 + 4 + 5 + 6 [n = 6, 6 addend.]
6 = -11 + ... + 0 + ... + 12. [n = 12, 12 addends.]
Therefore, any numbers that%26#039;s divisible by 1 %26amp; 2 can be x!
I hope this proves that this conjecture is true!
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