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# Understand the problem

Prove that the positive integers $$n$$ that cannot be written as a sum of $$r$$ consecutive positive integers, with $$r>1$$ ,are of the form $$n=2^l$$ for some $$l\ge 0$$.

##### Source of the problem

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 1.

Number Theory

8.5 out of 10

##### Suggested Book

Elementary Number Theory by David M. Burton

Do you really need a hint? Try it first!

Claim: Any positive integer $$n$$ can be written as $$n=2^k\cdot m$$ , where $$k\ge0$$ and $$m$$ is an odd positive integer.

To prove this claim use the fact : $$n=2^k\cdot p_1^{k_1}\cdot p_2^{k_2}\cdots p_i^{k_i}$$, where $$k$$ and all $$k_i$$ are non-negetive integers and all $$p_i$$ are odd primes.

The sum of (any) $$r$$ consecutive positive integers is given by,

$$(q+1)+(q+2)+(q+3)+\cdots+(q+r)$$

= $$q\cdot r+(1+2+3+\cdots+r)$$

= $$q\cdot r+\frac{r(r+1)}{2}$$.

Equating this sum to $$n$$ we get,

$$2^k\cdot m = q\cdot r+\frac{r(r+1)}{2}$$

Or, $$2^{k+1}\cdot m = 2q\cdot r+r(r+1)$$

Or, $$2^l\cdot m = r(2q+r+1)$$ , where $$l=k+1\ge1$$.

In particular if we take $$n=2^l$$ then $$m$$ is equal to 1.

Since both $$r$$ and $$(2q+r+1)$$ are greater than 1, so they can’t be equal to $$m$$ in this case. Again one of $$r, (2q+r+1)$$ is odd integer which implies the product $$r(2q+r+1)$$ can’t be equal to $$2^l$$.

$$\Rightarrow 2^l \neq r(2q+r+1)$$

$$\Rightarrow 2^l$$ can’t be expressed as the sum of $$r$$ consecutive positive integers with $$r>1$$ and $$l\ge 1$$.

Now, $$n=2^0=1$$ also can’t be written in the same manner when $$l=0$$. Therefore the positive integers $$n$$ that cannot be written as a sum of $$r$$ consecutive positive integers, with $$r>1$$ , are of the form $$n=2^l$$ for some $$l\ge 0$$.

To show that any number other than of the form $$2^l$$ is sum of consecutive integers:
Observe that for $$n=2^l.m$$, where $$m$$ is an odd number greater than 1, there are two cases:
1. $$m < 2^l$$
Select $$r$$ and $$q$$ such that $$r = m$$ and $$2q+r+1 = 2^l$$.

2. $$m > 2^l$$
Select $$r$$ and $$q$$ such that $$2q+r+1 = m$$ and $$r = 2^l$$.

(QED).

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