What-is-the-maximum-possible-value-of-k-for-which-2013-can-be-written-as-a-sum-of-k-consecutive-positive-integers-

Question Number 19700 by Tinkutara last updated on 14/Aug/17

What is the maximum possible value of

k for which 2013 can be written as a

sum of k consecutive positive integers ?

Answered by mrW1 last updated on 15/Aug/17

keep in mind : 2013 = 3 \times 11 \times 61

let a = the first number, a ⩾ 1

if k is even, i . e . k = 2 n

a = \frac{2013}{2 n} - \frac{2 n - 1}{2} = \frac{1}{2} (\frac{2013}{n} - 2 n + 1) ⩾ 1

\Rightarrow \frac{2013}{n} - 2 n ⩾ 1

\Rightarrow {2 n}^{2} + n - 2013 ⩽ 0

n ⩽ \frac{- 1 + \sqrt{1 + 8 \times 2013}}{4} \approx 31.5

since 2013 = 1 \times 3 \times 11 \times 61

\Rightarrow n = 1, 3, 11

\Rightarrow k = 2, 6, 22

if k is odd, i . e . k = 2 n + 1

a = \frac{2013}{2 n + 1} - n ⩾ 1

2013 - {2 n}^{2} - n ⩾ 2 n + 1

{2 n}^{2} + 3 n - 2012 ⩽ 0

n ⩽ \frac{- 3 + \sqrt{9 + 8 \times 2012}}{4} \approx 30.9

\Rightarrow 2 n + 1 ⩽ 62.9

since 2013 = 1 \times 3 \times 11 \times 61

\Rightarrow k = 2 n + 1 = 1, 3, 11, 33, 61

all possibilities are

k = 1, 2, 3, 6, 11, 22, 33, 61

k_{\max} = 61

Commented by Tinkutara last updated on 15/Aug/17

Thank you very much Sir!