In the set of consecutive odd numbers {1, 3, 5, ….., 57}, there is a number of \(k\) such that the sum of all the elements less than \(k\) is equal to the sum of all the elements greater than \(k\). Then, \(k\) equals.
41
The set has 29 odd numbers and their sum is \(29^{2}=841\).
Let \(k=2m-1\). Sum of odd numbers below \(k\) = \((m-1)^{2}\); sum above \(k\) = \(841-m^{2}\).
\((m-1)^{2}=841-m^{2} \Rightarrow 2m^{2}-2m-840=0 \Rightarrow m^{2}-m-420=0 \Rightarrow m=21\).
\(k=2(21)-1=41\). Check: \(20^{2}=400\) and \(841-441=400\).
Hence, 41.
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