Math Is Fun Forum

63lemon

Member

Registered: 2008-10-16

Posts: 0

Consecutive Numbers

can 16 or 32 or 64 be written as the sum of consecutive numbers?

fusilli_jerry89

Member

Registered: 2006-06-23

Posts: 86

Re: Consecutive Numbers

yes

fusilli_jerry89

Member

Registered: 2006-06-23

Posts: 86

Re: Consecutive Numbers

if u take half of an integer and keep adding half of that and half of that to infinite you will get the integer.

For example: 64 can be written 32+16+8+4+2+1+.5+.25+.125+....................etc.

This will equal 64 since the limit is 64.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Consecutive Numbers

There's a nice formula for the sum of an odd amount of consecutive numbers.
Say you wanted to find 5+6+7+8+9. By shifting 2 from the 9 to the 5, and 1 from the 8 to the 6, this becomes 7+7+7+7+7 = 5x7 = 35.

In its most general form, this method is:
(n-k)+(n-(k-1))+...(n-1)+n+(n+1)+...(n+(k-1))+(n+k).
= n+n+...+n+n [2k+1 times]
= n(2k+1)

So the useful point here is that for a number to be written as the sum of an odd amount of consecutive numbers, it needs to be a multiple of that odd amount. 16, 32 and 64 are all powers of 2 and so don't have any odd factors, so you can't do it that way.

Summing an even amount of consecutive numbers isn't as nice, but we can still narrow down the search.
Using O and E to mean a general odd and even number respectively, the sum of two consecutive numbers will always be O+E or E+O. Either way, the answer will be odd, so 16, 32 and 64 can't be done this way.

O+E+O+E is even though. In fact, it's fairly easy to extend this and see that adding an amount of consecutive numbers that is a multiple of 4 gets you an even number, and adding an amount of consecutive numbers that is even but not a multiple of 4 gets you an odd one.

So we want to consider sums that contain either 4 or 8 terms.
(The lowest sum containing 12 terms is 1+2+...+11+12 = 78, which is too high, and sums containing even more terms will be even higher)

Let's try it with 4 terms first (eg. 1+2+3+4).
In general form, a sum of 4 terms looks like n + (n+1) + (n+2) + (n+3).
This simplifies to 4n + 6. However, 16, 32 and 64 are all multiples of 4 and 4n+6 never is (because 4n is and 6 isn't), so no value of n will work here.

Our only option left is an 8-term sum.
The general form is similar to above: n + (n+1) + ... + (n+7).
Simplifying gives 8n + 28, but now we run into the same problem as before.
16, 32 and 64 are all multiples of 8, but 8n+28 never is, so no value of n gives a solution here either.

So, the answer is that none of 16, 32 and 64 can be written as the sum of consecutive numbers.

---

Why did the vector cross the road?
It wanted to be normal.

sce1912

Member

Registered: 2008-10-08

Posts: 5

Re: Consecutive Numbers

Hi all,

Found a nice generalisation of this problem stating that the sum of 2 or more consecutive integers cannot be a power of 2

Proof:
Let n, m and t be non-negative integers
Using that result that 1 + 2 + ... + n = 1/2n(n+1)

We have
(m+1) + (m+2) + ... + n = 1/2n(n+1) - 1/2m(m+1) = 1/2[n^2 + n - m^2 - m]  = 1/2[(n+m)(n-m) + (n-m)] = 1/2(n-m)(m+n+1)

Now suppose
1/2(n-m)(m+n+1) = 2^t
Then  (n-m)(m+n+1) =2^(t+1)
but one of (n-m) and (n+m+1) must be odd as their sum is odd and the only odd factor of 2^(t+1) is 1
This means either n = m = 0 (no sum at all) or n = m+ 1 (which means the sum only contains 1 number)
This contradiction proves the result for non-negative integers.

Including negative integers doesn't help matters since either the sum will be negative or we can cancel out all negative numbers because (-n) + n = 0

Last edited by sce1912 (2008-10-17 05:26:33)