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).
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