One day I was playing with my calculator, I pushed \(6^5\) and then it came out 7776. I found it very interesting, because except the units digit, all the other digits are 7!

Then a problem came up in my mind:

How many integers in the form \(a^b\) where \(a\) and \(b\) are positive integers greater than 1 which except the units digits, the other digits are all the same?

This problem is equivalent to finding integer solutions that satisfy \[a^b=\overline{\underbrace{mmm\ldots mm}_{x~m\text{'s}}n}\] Here, \(x>1\).

There is another number I found: \(21^2=441\)

Any ideas?

If you have a solution, comment and list out the steps.

## Comments

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TopNewestA computer search shows that for numbers up to \(60\) digits long, the only ones that have this property are \(225, 441, 7776\). If there exists another, it's going to be longer than \(60\) digits. – Michael Mendrin · 2 years, 10 months ago

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