2. Find all three-digit numbers that are equal to the third power of the sum of their digits.
Solution. Let $\overline{a b c}$ be the desired number. Then $\overline{a b c}=(a+b+c)^{3}$. Since $\overline{a b c}$ is a three-digit number, the equation will hold if $5 \leq a+b+c \leq 9$. If $a+b+c=5$ then $(a+b+c)^{3}$ ends in 5. However, then $\overline{a b c}$ also ends in 5, from which it follows that $a+b+c>5$, which is a contradiction. Similarly, if $a+b+c=6$ or $a+b+c=9$ then $(a+b+c)^{3}$ ends in 6 or 9, and again a contradiction is obtained. It remains to check the cases $a+b+c=7$ and $a+b+c=8$. If $a+b+c=7$, then $\overline{a b c}=7^{3}=343$, but $3+4+3 \neq 7$. The remaining case is $a+b+c=8$. Then $\overline{a b c}=8^{3}=512$ and $5+1+2=8$. The desired number is 512.