A, B, P are distinct digits.

AA, AAA, ABB are concatenated numbers.



$$(AA + ABB)^P-(ABB)^P=(AAA - A) ^P+(B+B^B*B)^P+(B^B)^P$$

Don’t be afraid to make reasonable assumptions to quickly deduce A and B.

  • $\begingroup$ Thx..looks much better and distinguishes both equations $\endgroup$
    – Uvc
    Jun 7, 2019 at 11:29

1 Answer 1


I believe A, B, P are

1, 3, 5

Some deductions I came to before brute forcing

A must be a value between 1 and 4. If A is zero, then B and P are equal which violates the distinct digits, and if A is 5 or greater, than P must be 10 or more. Due to the nature of the exponents I didn't think A would be particularly large so I simply plugged in 1 for A and worked from there. I tried B = 2 which didn't work and then B = 3 which gave me a value of 20301568331 on both sides of the second equation.

  • $\begingroup$ You can also notice that on the left hand side of the long equality you can factor out an A^2p from every term, but only the first term on the right depends on A, so A must be 1. $\endgroup$
    – JProblems
    Jun 7, 2019 at 13:24
  • $\begingroup$ @jproblems good observation $\endgroup$
    – Uvc
    Jun 7, 2019 at 14:12
  • $\begingroup$ Right logic tames even higher powers leading to quick result! $\endgroup$
    – Uvc
    Jun 7, 2019 at 14:17

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