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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.

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

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    $\begingroup$ Thx for the edit..looks better $\endgroup$ – Uvc Jun 23 at 9:26
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    $\begingroup$ You're welcome. $\endgroup$ – Ak19 Jun 23 at 9:26
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    $\begingroup$ 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. $\endgroup$ – TheSimpliFire Jun 23 at 9:46
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    $\begingroup$ Sure..statement regarding prime relationship gives valuable clue...if terms are rearranged..this is the first number that fails it.. $\endgroup$ – Uvc Jun 23 at 9:49
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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.

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  • $\begingroup$ Prof. Roman congratulates you for your fast thinking and awards you extra credit of AAA points.. $\endgroup$ – Uvc Jun 23 at 9:41
  • $\begingroup$ @Uvc do you mean B? how can you award AAA points??? $\endgroup$ – Omega Krypton Jun 23 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 at 9:45
  • $\begingroup$ AAA is 3 points? $\endgroup$ – JS1 Jun 23 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 at 10:26

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