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--- pow:problem1f24 [2024/09/10 04:46] – mazur
+++ pow:problem1f24 [2024/09/10 04:47] (current) – mazur
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+<box 85% round orange|Problem 1 (due on Monday, September 9) >
+Find all natural numbers $n>1$ such that $2!+3!+\ldots+n!$ is a cube of an integer.
+</box>
+The problem was solved by Sasha Aksenchuk, Prof. Vladislav Kargin, Josiah Moltz, and Mithun Padinhare Veettil.
+The only solution is $n=3$. All solutions received and our in-house solution are based on the observation
+that a cube of an integers must yield a remainder 0, 1, or 6 when divided by 7. For a detailed solution see the following link {{:pow:2024fproblem1.pdf|Solution}}.