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Professor Roman gives unusual math quiz ahead of

This is not math. This is war!Professor Delvershys calculationsRepresentation of Mo-roman numeralsSimple Math Problem #1Simple Math Problem #3A Simple Math PuzzleThe Pebbles QuizGreco-Roman egalitarianismRoman Numerals Equation 2Math quiz from a book

11

$begingroup$

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.

$$bbox[5px,border:2px solid red]AA^BA-big(BBAAAcdot CBBBABbig) = A$$

mathematics lateral-thinking calculation-puzzle no-computers

share|improve this question

edited Jun 23 at 22:29

Uvc

asked Jun 23 at 8:55

UvcUvc

2,60355 silver badges2929 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Thx for the edit..looks better
  $endgroup$
  – Uvc
  Jun 23 at 9:26
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome.
  $endgroup$
  – Ak19
  Jun 23 at 9:26
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Comment proving that actual 'digits' are impossible. The term $BBAAAcdot CBBBAB$ 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 $$AA^A-(AAAcdot C000A0)equiv(11A)^A-111Acdot(100000C+10A)=A$$ 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
  

  
  
  
  


• 
  
  
  

  
  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 at 9:49
  

  
  
  
  

add a comment | 

11

$begingroup$

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.

$$bbox[5px,border:2px solid red]AA^BA-big(BBAAAcdot CBBBABbig) = A$$

mathematics lateral-thinking calculation-puzzle no-computers

share|improve this question

edited Jun 23 at 22:29

Uvc

asked Jun 23 at 8:55

UvcUvc

2,60355 silver badges2929 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Thx for the edit..looks better
  $endgroup$
  – Uvc
  Jun 23 at 9:26
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome.
  $endgroup$
  – Ak19
  Jun 23 at 9:26
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Comment proving that actual 'digits' are impossible. The term $BBAAAcdot CBBBAB$ 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 $$AA^A-(AAAcdot C000A0)equiv(11A)^A-111Acdot(100000C+10A)=A$$ 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
  

  
  
  
  


• 
  
  
  

  
  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 at 9:49
  

  
  
  
  

add a comment | 

$begingroup$

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.

$$bbox[5px,border:2px solid red]AA^BA-big(BBAAAcdot CBBBABbig) = A$$

mathematics lateral-thinking calculation-puzzle no-computers

share|improve this question

edited Jun 23 at 22:29

Uvc

asked Jun 23 at 8:55

UvcUvc

2,60355 silver badges2929 bronze badges

$endgroup$

share|improve this question

edited Jun 23 at 22:29

Uvc

asked Jun 23 at 8:55

UvcUvc

2,60355 silver badges2929 bronze badges

share|improve this question

edited Jun 23 at 22:29

Uvc

asked Jun 23 at 8:55

UvcUvc

2,60355 silver badges2929 bronze badges

asked Jun 23 at 8:55

UvcUvc

2,60355 silver badges2929 bronze badges

asked Jun 23 at 8:55

UvcUvc

2,60355 silver badges2929 bronze badges

1 Answer
1

active

oldest

votes

10

$begingroup$

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.

share|improve this answer

edited Jun 23 at 9:42

answered Jun 23 at 9:38

JS1JS1

4,5621717 silver badges2626 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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10

$begingroup$

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.

share|improve this answer

edited Jun 23 at 9:42

answered Jun 23 at 9:38

JS1JS1

4,5621717 silver badges2626 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

10

$begingroup$

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.

share|improve this answer

edited Jun 23 at 9:42

answered Jun 23 at 9:38

JS1JS1

4,5621717 silver badges2626 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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.

share|improve this answer

edited Jun 23 at 9:42

answered Jun 23 at 9:38

JS1JS1

4,5621717 silver badges2626 bronze badges

$endgroup$

share|improve this answer

edited Jun 23 at 9:42

answered Jun 23 at 9:38

JS1JS1

4,5621717 silver badges2626 bronze badges

answered Jun 23 at 9:38

JS1JS1

4,5621717 silver badges2626 bronze badges

answered Jun 23 at 9:38

JS1JS1

4,5621717 silver badges2626 bronze badges