MHB Find Number b/w 1000-2000: Impossible Sum of Consec. Nums

[kaliprasad]

Find the number between 1000 and 2000 that cannot be expressed as sum of (that is >1) consecutive numbers.( To give example of sum of consecutive numbers

101 = 50 + 51

162 = 53 + 54 + 55 )

and show that it cannot be done

 

[Opalg]

Find the number between 1000 and 2000 that cannot be expressed as sum of (that is >1) consecutive numbers.( To give example of sum of consecutive numbers

101 = 50 + 51

162 = 53 + 54 + 55 )

and show that it cannot be done

[sp]Suppose that $n$ is a multiple of an odd number, say $n = k(2p+1)$. Then $$n = \sum_{r=-p}^p (k+r)$$, so that $n$ is a sum of consecutive integers.

[Note: In that sum, some of the terms may be negative. But if so, those terms will cancel with some of the positive terms, leaving an expression for $n$ as a sum of consecutive positive integers. For example, if $k = 2$ and $p=3$ then $n = k(2p+1) = 14$, and the above formula gives $14 = (-1) + 0 + 1 + 2 + 3 + 4 + 5$. After cancelling the $-1$ with the $+1$, that becomes $14 = 2+3+4+5$.]

So if number cannot be expressed as a sum of that form then it can have no odd divisors at all and must therefore be a power of 2. If the number lies between 1000 and 2000 then it has to be 1024.

To see that a power of 2 cannot be a sum of that form, notice that if such a sum has an odd number of terms, say $2p+1$ terms with the middle term being $k$, then the sum is $k(2p+1)$ (as above) and therefore has an odd factor. If the sum has an even number of terms, say $2k$ terms starting with $m$, then the sum is $$\sum_{r=0}^{2k-1}(m+r) = k(2m+2k-1),$$ which again has an odd factor, $2(m+k)-1.$ Therefore a power of 2 can never be a sum of consecutive integers.[/sp]

 

[kaliprasad]

[sp]Suppose that $n$ is a multiple of an odd number, say $n = k(2p+1)$. Then $$n = \sum_{r=-p}^p (k+r)$$, so that $n$ is a sum of consecutive integers.

[Note: In that sum, some of the terms may be negative. But if so, those terms will cancel with some of the positive terms, leaving an expression for $n$ as a sum of consecutive positive integers. For example, if $k = 2$ and $p=3$ then $n = k(2p+1) = 14$, and the above formula gives $14 = (-1) + 0 + 1 + 2 + 3 + 4 + 5$. After cancelling the $-1$ with the $+1$, that becomes $14 = 2+3+4+5$.]

So if number cannot be expressed as a sum of that form then it can have no odd divisors at all and must therefore be a power of 2. If the number lies between 1000 and 2000 then it has to be 1024.

To see that a power of 2 cannot be a sum of that form, notice that if such a sum has an odd number of terms, say $2p+1$ terms with the middle term being $k$, then the sum is $k(2p+1)$ (as above) and therefore has an odd factor. If the sum has an even number of terms, say $2k$ terms starting with $m$, then the sum is $$\sum_{r=0}^{2k-1}(m+r) = k(2m+2k-1),$$ which again has an odd factor, $2(m+k)-1.$ Therefore a power of 2 can never be a sum of consecutive integers.[/sp]

Good solution above by opalg.

my solution

Spoiler

for it to be a sum of consecutive numbers it must have an odd factor

Let N be a number that can be represented as a sum of k > 1 consecutive integers, starting with, let's say, a+1, summing the arithmetic progression, we get:
$N = (a+1) + (a+2) + (a+3) + . . + (a+k) = \frac(ka + k(k + 1)){2}$

Hence $2N = 2ka + k(k + 1) = k(2a + k + 1)$
Now if k is odd, N is divisible by k

If k is even, then 2a + k + 1 are odd and again N would have at least one odd factor.>1

Spoiler

Then I need to show that if it has a odd factor > 1 this can be expressed as sum of of consecutive number

2n + 1 = n + (n+1)

Any even number is a power of 2 or can be expressed as power of 2 multiplied by an odd number >1

Say it is $2^p(2n+1)$ where p is highest power of 2 which is a factor.

Now 2n+1 can be written as n+ (n+1) of course n > 1

Now we have 2 cases $2^p \lt n$ or $2^p \gt n$

Case 1) $2^p < n$

This can occur $2^p$ times but there shall be $2^p$ copies of n and $2^p$ copies on n + 1

In the kth copy subtract k-1 from n and add k-1 to n+ 1

So we get the values $n-(2^p-1)$ to $n+ 2^p$

The sum shall be $2^p ( n + 1 +n)$

And all numbers are consecutive ( 1st number >= 1) and positiveCase 2) $2^p > n$

We take 2n+1 copies of $2^p$

From the middle value at a distance of k we subtract k on from the left element and add k on the right element

So we have consecutive numbers $2^p-n$ to $2^p+ n$

Spoiler

Hence for it not to be sum of consecutive numbers it has to be power of 2 that is 1024