Prove $n^3$ has the form $9k$ or $9k + 1$ or $9k + 8$ for some integer $k$.
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Any number can be written as one of $3k1$, $3k$ or $3k + 1$ for some $k$. It's obvious what happens in the $3k$ case. Otherwise, $$ (3k \pm 1)^3 = 27k^3 \pm 3 \cdot 9k^2 + 3 \cdot 3k \pm 1 \equiv \pm 1 \pmod{9}. $$
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AH Keng
Updated on December 26, 2022Comments

AH Keng 11 months
Let $n$ be an integer. Prove $n^3$ has the form $9k$ or $9k + 1$ or $9k + 8$ for some integer $k$.

AH Keng about 6 yearsAnyone can help?
