Henry Ernest Dudeney/Puzzles and Curious Problems/122 - A Question of Cubes/Solution
Jump to navigation
Jump to search
Puzzles and Curious Problems by Henry Ernest Dudeney: $122$
- A Question of Cubes
- From Sum of Sequence of Cubes, the cubes of successive numbers, starting from $1$, sum to a square number.
- Thus the cubes of $1$, $2$, $3$ (that is, $1$, $8$, $27$), add to $36$, which is the square of $6$.
- If you are forbidden to use the $1$, the lowest answer is the cubes of $23$, $24$ and $25$, which together equal $204^2$.
- What is the next lowest number, using more than three consecutive cubes and as many more as you like, but excluding $1$?
Solution
- $\ds \sum_{n \mathop = 14}^{25} n^3 = 14^3 + 15^3 + 16^3 + \cdots + 25^3 = 97 \, 344 = 312^2$
The next one after that is:
- $\ds \sum_{n \mathop = 25}^{29} n^3 = 25^3 + 26^3 + 27^3 + 28^3 + 29^3 = 99 \, 225 = 315^2$
Proof
This theorem requires a proof. In particular: I'm surprised we haven't got this in already. I would have expected David Wells to be all over this, but it seems he missed it. Unless he did, we do have it, and I'm missing it now. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Sources
- 1932: Henry Ernest Dudeney: Puzzles and Curious Problems ... (previous) ... (next): Solutions: $122$. -- A Question of Cubes
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $195$. A Question of Cubes