Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to find a three digit number which,when reversed, becomes equal to $17$ times the square of it's cube root?

If we assume that the three digit number is of the form $100x+10y+z$,where $x \in [1,9]$ and $y,z \in [0,9]$.It seems to me that we have to solve $(x,y,z)$ from the equation $$100z+10y+x = 17 \times (100x+10y+z)^\frac{2}{3}$$

but I just can't see what to do next,any ideas?

share|cite|improve this question
There aren't that many 3-digit numbers that have cube roots :) – Srivatsan Sep 26 '11 at 2:50
@Srivatsan Narayanan :Indeed only five:$125,216,343,512$ and $729$ – Quixotic Sep 26 '11 at 2:58
up vote 2 down vote accepted

You just need to iterate through the perfect cubes smaller than $1000$, and check which of them satisfies the condition. (The answer is $216$.)

Justification: The given condition is that $N' = 17N^{2/3}$ where $N'$ is the "reverse" of $N$. So $N^{2/3}$ is rational, and this is possible iff $N$ is a perfect cube. (More generally, if $N, a, b$ are positive integers such that $\gcd(a,b)=1$ and $N^{a/b}$ is rational, then $N$ is a perfect $b$-th power. This statement can be proved via the fundamental theorem of arithmetic.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.