Let $\overline{abc}$ be $3$-digits number. What is the maximal value of the expression $\overline{abc}-(a^3+b^3+c^3)?$
Computer gives the answer $396$ for the number $620$ but I have no idea how to do it by hand.
What is the maximal value of the expression $\overline{abc}-(a^3+b^3+c^3)?$
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algebra-precalculus
2 Answers
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You can write the whole thing as $100a + 10b + c - a^3 - b^3 - c^3$.
Rearranging the terms gives $(100a - a^3) + (10b - b^3) + (c - c^3)$, which you can solve one summand at a time:
You maximize $100a - a^3$ for $a = 6$.
You maximize $10b - b^3$ for $b = 2$.
You maximize $c - c^3$ for $c = 0$ or $1$.
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You can write $\overline{abc} = 100a + 10 b + c$. Then $$\overline{abc} - (a^3 + b^3 + c^3) = a(100-a^2) + b(10 - b^2) + c(1-c^2).$$ You just need to find the digits $a,b,c$ that maximize each term individually.