Easier than It Looks

There are four three digit natural numbers that satisfy the condition that each the sum of the cubes of its digits.

Here are the three of these numbers:

\[ \begin{eqnarray} 153&=&1^3+5^3+3^3 \\ 370&=&3^3+7^3+0^3 \\ 407&=&4^3+0^3+7^3 \\ \end{eqnarray} \]

Can you find the remaining missing number?

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