His usual Monday Morning 8am class. This is for extra AAA credits.

$A$, $B$, $C$ are distinct digits.

$AA$, $BA$, $BBAAA$, $CBBBAB$ are distinct numbers.

Please deduce these with concise reasoning from the given relation:

Rearrangement of the terms in the Equation gives interesting Prime Relationship.

$$\bbox[5px,border:2px solid red]{AA^{BA}-\big(BBAAA\cdot CBBBAB\big) = A}$$

  • 2
    $\begingroup$ Thx for the edit..looks better $\endgroup$
    – Uvc
    Jun 23, 2019 at 9:26
  • 3
    $\begingroup$ You're welcome. $\endgroup$
    – 19aksh
    Jun 23, 2019 at 9:26
  • 2
    $\begingroup$ Comment proving that actual 'digits' are impossible. The term $BBAAA\cdot CBBBAB$ has a maximum order of magnitude of $(4+5)+1=10$ where the $1$ is due to carrying. However, if $B>0$, $AA^{BA}$ will have order of magnitude of at least $11$ since its minimum value is $11^{11}$. We cannot have $A=0$ based on the assumption that $0^0$ is not allowed. Thus $B=0$. This leaves us with the expression $$AA^A-(AAA\cdot C000A0)\equiv(11A)^A-111A\cdot(100000C+10A)=A$$ and we can see that $A^A$ must have the same last digit as $A$, forcing $A=5,6,9$. Checking each value proves their impossibility. $\endgroup$ Jun 23, 2019 at 9:46
  • 1
    $\begingroup$ Sure..statement regarding prime relationship gives valuable clue...if terms are rearranged..this is the first number that fails it.. $\endgroup$
    – Uvc
    Jun 23, 2019 at 9:49

1 Answer 1


The answer uses Roman numerals, as this is a Prof. Roman puzzle.

A = I
B = X
C = L

This gives the equation:

$2^{11} - 23*89 = 1$

Reasoning: after guessing roman numerals were involved, the pattern CBBBAB forces AB to be IX or XC or CM etc. This is because BAB forces A to be a one type digit, but BBB means that B can't be a five type digit because something like VVVIV doesn't exist (it would be XVIV). Taking the simplest case of A=I B=X gives $2^{11} - 23 * (C+39) = 1$, which means that C must be 50, or L.

  • $\begingroup$ Prof. Roman congratulates you for your fast thinking and awards you extra credit of AAA points.. $\endgroup$
    – Uvc
    Jun 23, 2019 at 9:41
  • $\begingroup$ @Uvc do you mean B? how can you award AAA points??? $\endgroup$ Jun 23, 2019 at 9:42
  • $\begingroup$ I mean B..thx..regarding lateral thinking..probably debatable? I wanted to hint that it is not regular number substitution for the letters $\endgroup$
    – Uvc
    Jun 23, 2019 at 9:45
  • $\begingroup$ AAA is 3 points? $\endgroup$
    – JS1
    Jun 23, 2019 at 9:52
  • $\begingroup$ After solving..it is 111 points...in the puzzle it should mean triple A credits rather than points strictly to avoid confusion $\endgroup$
    – Uvc
    Jun 23, 2019 at 10:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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