This post is going to be a pretty short one. Now that I have tasted blood again, I managed to “solve” another Project Euler problem just before hitting the hay last night. I’m still working on the supposed to be easy ones though.

I put “solved” in quotation marks due to the massive sledgehammer which I used for arriving at the solution…a simple for-loop. Does 1 satisfy the given definition? What about 2? And 3? Bang! Bang! Bang!

Using a dumb loop was the first thought that occurred to me on how I could solve this challenge and I didn’t get any smarter in the end.

I started by consulting Wolfram MathWorld to find out more about this numerical phenomenon. Just to discover Narcissistic Numbers. Which is NOT the solution to this given problem.

An

n-digit number that is the sum of thenth powers of its digits is called ann-narcissistic number.

Below this definition MathWorld displays a nice little table of all the numbers that satisfy this definition up to *n* equals 39.

That was easy. For *n* equals 5 the numbers I was looking for are 54748, 92727, 93084.

Unfortunately…

These are not the numbers you’re looking for.

Obi -Wan Kenobi

Or to be a bit more precise, these are not the only numbers required to reach the target sum.

After several failed attempts to submit 240559 (54748 + 92727 + 93084) I came to the conclusion that most likely I am not smarter than Project Euler and started thinking about what I was missing here.

If you read the problem description carefully, it becomes clear that your search is not limited to ** n**-narcissistic numbers only. For example 4150 is just fine as well:

4150: 4^{5} + 1^{5} + 5^{5} + 0^{5} = 4150.

The number doesn’t have to be precisely *n*-digits long. I hope this helps.

Now I have to read the forum posts as usual. And then move on to the next problem on my list, Coin sums.