In the set of consecutive odd numbers $\{1, 3, 5, \ldots, 57\}$, there is a number $k$ 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?
Show Hint
For consecutive odd numbers, remember: sum of first \(n\) terms = \(n^2\). This simplifies balance-sum problems significantly.
Approach (guess-and-verify the balance using $n^2$ chunks): Each option is one of the listed odd numbers; for each, the part below it is a block of the first few odds, whose sum is a perfect square. Test which split is symmetric.
Step 1: Total of all $29$ odds is $29^2 = 841$. For $k$ to balance, the numbers strictly below $k$ must sum to $\tfrac{841 - k}{2}$ (half of what's left after removing $k$). So I need $\tfrac{841 - k}{2}$ to equal the sum of the odds below $k$.
Step 2: Try $k = 41$ (the $21$st odd number). Numbers below it are the first $20$ odds, summing to $20^2 = 400$. Required value: $\tfrac{841 - 41}{2} = \tfrac{800}{2} = 400$. Match.
Step 3 (why the others fail): For $k = 39$ ($20$th odd), below-sum $= 19^2 = 361$, but $\tfrac{841-39}{2} = 401$ — no. For $k = 43$ ($22$nd odd), below-sum $= 21^2 = 441$, but $\tfrac{841-43}{2} = 399$ — no. For $k = 37$ ($19$th odd), below-sum $= 18^2 = 324$ vs $\tfrac{841-37}{2}=402$ — no.
Answer: Only $k = 41$ splits the set evenly into $400$ and $400$.