2010 UNCO Math Contest II Problems/Problem 1

Problem

Find a $3$-digit integer less than $200$ where each digit is odd and the sum of the cubes of the digits is the original number.

Solution

We have a three digit number that can be represented as $1bc$. We know that $b$ and $c$ are odd, and that $b^3 + c^3 + 1 = 100 + 10b + c$. This means that both $b$ and $c$ are within the set ${1,3,5,7,9}$. We also notice that since $7^3$ and $9^3$ are both far greater than $200$, they can not be our digits, so the remaining set is ${1,3,5}$. Looking at only the last digits of $1^3 = 1, 3^3 = 27, 5^3 = 125$, we see that only when $125$ and $27$ are together can we get a last digit of $5$ or $3$, which we see is $125+27+1 = 153$. Upon checking, $\boxed{\textbf{153}}$ is a working solution.

See also

2010 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions