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# Powers of 2 – I.S.I. Entrance 2019 Subjective Problem 1

## Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"] 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 $l\ge 0$.

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

[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.4" open="off"]Number Theory

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8.5 out of 10

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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.

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The sum of (any) $r$ consecutive positive integers is given by, $(q+1)+(q+2)+(q+3)+\cdots+(q+r)$ =   $q\cdot r+(1+2+3+\cdots+r)$ =   $q\cdot r+\frac{r(r+1)}{2}$.

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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 $l=k+1\ge1$.

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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).

## Connected Program at Cheenta

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

## Similar Problem

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