Fun Maths Problem

[Roflcopter]

Find, with proof, the set of natural numbers that cannot be written as the sum of two or more consecutive non-negative integers. Natural numbers are just positive whole numbers.

Here are some examples:

3 = 1+2
5 = 2+3
6 = 1+2+3
11 = 5+6

Bonus marks if you can find a simple method to find the actual consecutive numbers.

I will post the answer if there's no sign of it getting solved.

[Enam]

For the bonus marks

It works for all natural numbers, apart from the number 1 and any number which is a power of 2.

Power's of 2 have no odd factors, it will be impossible to make 2 or more consecutive numbers add up to any natural number without the use of an odd number.

The formula to work out the numbers which can be written as the sum of two or more consecutive non-negative integers (the nth term).
2n + 1 = n + (n + 1)

[#M|A#Wolf]

I have no idea what you just said.

[Roflcopter]

So Enam here provides a proof that all numbers divisible by an odd prime must be writeable in this way.

For odd numbers 2n+1 = n + (n+1).

For even numbers n with an odd prime factor p, n/p is an integer which we can centre a sequence of consecutive numbers around. That is n = sum(j+i, i=0..p-1) with j = n/p - (p-1)/2.

But it remains to prove that 2^m cannot be written in this way (for m>0).

[Felix-The-Ghost]

But it remains to prove that 2^m cannot be written in this way (for m>0).

Two is the only even prime number. So any powers of it cannot contain an odd prime factor, which are needed to be written that way.

[Roflcopter]

Two is the only even prime number. So any powers of it cannot contain an odd prime factor, which are needed to be written that way.

That begs the question why we need an odd prime? I mean it's true (excluding the case 1=0+1), but you're making too big a leap from "no odd prime factors" to "can't be written as we want".

[Felix-The-Ghost]

So Enam here provides a proof that all numbers divisible by an odd prime must be writeable in this way.

You said it yourself, but I guess you didn't explicitly say that is the only case where they are writable in that way (needs a prime factor)

[Roflcopter]

The implication works only one way. All natural numbers divisible by an odd prime are writeable in the desired form. But this doesn't imply that all other numbers aren't.

[Felix-The-Ghost]

The sum of a sequence of consecutive positive integers can be expressed as

sum = (number of elements * average of elements)

because the elements are evenly spaced numbers.

For example in the sequence [2, 3, 4, 5, 6]

5*4 = 20

In the example the sequence has an odd number of elements and thus the sum is divisible by an odd integer (in this case five)

For sequences with an even number of elements, you have a situation, where the average is not an integer, for example the sequence [2, 3, 4, 5, 6, 7, 8, 9]

8 * (5.5) = S

to make the equation easier you can divide the first factor by two and multiply the second by two because this will have the same result.

4 * 11

This makes the right factor odd, meaning the sum of such sequences are always divisible by an odd integer. Therefore the only positive numbers not expressible by such sequences are those that have only even factors. Two is the only even prime number, meaning powers of it are not divisible by odd integers, meaning that number cannot be expressed as the sum of sequential numbers.

[Marti]

2+2= fish,
3+3= eight,
7+7= triangle

'nuff said.

[MykeGregory]

inB4 asian jokes.
Seriously though, I wont even attempt these maths problems. NEVER have been good at maths.

[Xenon]

maths in a ac forum
disgusting

[Roflcopter]

...

Spot on!

[Andrez]

maths
disgusting

Maybe you really meant this!

[Link]

maths in a ac forum
disgusting

an* thread*
Ok, I totally agree with this.

[Marti]

maths in a ac forum
disgusting

an* thread*

gj failing. it should be "a thread" or "an ac forum"

[Felix-The-Ghost]

Bonus marks if you can find a simple method to find the actual consecutive numbers.

I guess we didn't explicitly do this:

Say you want to find a sequence of x consecutive numbers that total up to y.

If x is an odd number, y must be divisible by x for there to be a possible solution for x numbers.

e.g. find out which 3 numbers can make 21.

21/3 = 7 -- then you add an equal amount of numbers on both sides of this number until you have a sequence of x numbers and there's your answer.

6...7...8 (6+7+8=21)

If x is even, it's a little different. The quotient of y/x must be exactly between two integers. You can double y temporarily and half the quotient to see if it is between two integers (basically 2y/x must be an odd number)

e.g. find out which 4 numbers can make 14.

14/4 = 3.5 OR 14*2(28) / 4 (7) / 2 (3.5)

The middle numbers of your sequence will be Q - 0.5, Q + 0.5, where Q is the quotient to the above problem (when it is exactly between two integers)

3...4

Add numbers to both sides until you have a sequence of x numbers:

2...3...4...5 (2+3+4+5=14)

[FleshyPod]

Maths can be funny - explain that a+b=a in Boolean alghebra

[#M|A#Wolf]

b = 0? xD

[FleshyPod]

No,
a+b=a+(b+ <negated b>)=a

[Oracle]

So basically there was no b to begin with? \:D