I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 1.
8.5 out of 10
[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.3.4" open="off"]Elementary Number Theory by David M. Burton[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="4.3.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]
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.
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]The sum of (any) \(r\) consecutive positive integers is given by, \((q+1)+(q+2)+(q+3)+\cdots+(q+r)\) = \(q\cdot r+(1+2+3+\cdots+r)\) = \(q\cdot r+\frac{r(r+1)}{2}\).
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]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 \(l=k+1\ge1\).
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]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).
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.
[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/isicmientrance/" button_text="Learn More" button_alignment="center" _builder_version="3.22.4" custom_button="on" button_text_color="#ffffff" button_bg_color="#e02b20" button_border_color="#e02b20" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]
Very Good
Thank you. Keep doing great mathematics.
The proof is not yet complete.There may be such n which can't be expressed as the sum of consecutive integers but also is not of the form 2^l. To disprove it we have to show that any number other than of the form 2^l is some of consecutive integers. How can we prove this?
Thanks, Alapan for your valuable suggestion. We have made the necessary changes. Keep doing great Mathematics. Hope it helps now.