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Resolve this Fibonacci Relationship

Find the values of U, V, C based on the given relationship…useful for upcoming puzzlesFind this Unique UVC Palindrome ( ignoring signs and decimal) from Given Fractional RelationshipLady Luck Powers Up Every Member to Sum Upto Non-Prime Number. Who am I?UVC wants to give you a Helping Hand in Solving these Unique Set of Pan digital Fractional-Decimal RelationsFrom the given Square - Factorial Relationship, deduce the unknownsPlease figure out this Pan digital PrincePan Digital Lucky Seven wants you to figure out all the digitsA Lollipop with RootsPerfect Powered Relations - Please Figure them outResolve these Highly Narcissistic Relations

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

12

$begingroup$

$Given$:

$A$, $B$, $C$, $E$, $F$ are distinct digits varying from $1$ to $9$.

$A$ is a Fibonacci number.

$BB$, $BC$, $EF$ are concatenated Numbers.

$Relationship$:

$(A*BB)*(BC)^2$ = $(EF)^2- B$

Deduce all the Digits.

mathematics logical-deduction knowledge no-computers

share|improve this question

asked Jun 26 at 9:52

UvcUvc

2,69855 silver badges3131 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  are BB, BC, EF fib no.s?
  $endgroup$
  – Omega Krypton
  Jun 26 at 9:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Enough info is given to resolve
  $endgroup$
  – Uvc
  Jun 26 at 9:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do give the deductive reasoning which is very simple and straightforward.
  $endgroup$
  – Uvc
  Jun 26 at 10:19
  

  
  
  
  

add a comment | 

12

$begingroup$

$Given$:

$A$, $B$, $C$, $E$, $F$ are distinct digits varying from $1$ to $9$.

$A$ is a Fibonacci number.

$BB$, $BC$, $EF$ are concatenated Numbers.

$Relationship$:

$(A*BB)*(BC)^2$ = $(EF)^2- B$

Deduce all the Digits.

mathematics logical-deduction knowledge no-computers

share|improve this question

asked Jun 26 at 9:52

UvcUvc

2,69855 silver badges3131 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  are BB, BC, EF fib no.s?
  $endgroup$
  – Omega Krypton
  Jun 26 at 9:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Enough info is given to resolve
  $endgroup$
  – Uvc
  Jun 26 at 9:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do give the deductive reasoning which is very simple and straightforward.
  $endgroup$
  – Uvc
  Jun 26 at 10:19
  

  
  
  
  

add a comment | 

$begingroup$

$Given$:

$A$, $B$, $C$, $E$, $F$ are distinct digits varying from $1$ to $9$.

$A$ is a Fibonacci number.

$BB$, $BC$, $EF$ are concatenated Numbers.

$Relationship$:

$(A*BB)*(BC)^2$ = $(EF)^2- B$

Deduce all the Digits.

mathematics logical-deduction knowledge no-computers

share|improve this question

asked Jun 26 at 9:52

UvcUvc

2,69855 silver badges3131 bronze badges

$endgroup$

share|improve this question

asked Jun 26 at 9:52

UvcUvc

2,69855 silver badges3131 bronze badges

share|improve this question

asked Jun 26 at 9:52

UvcUvc

2,69855 silver badges3131 bronze badges

asked Jun 26 at 9:52

UvcUvc

2,69855 silver badges3131 bronze badges

asked Jun 26 at 9:52

UvcUvc

2,69855 silver badges3131 bronze badges

1 Answer
1

active

oldest

votes

12

$begingroup$

Answer

$$(A,B,C,E,F) = (5,1,2,8,9)$$

Reasoning

$$A * BB geq 22$$ $$ (EF)^2 - B leq 98^2-1 = 9603$$ $$ Rightarrow (BC)^2 leq frac960322 = frac8732 = 436 frac12$$ $$ Rightarrow BC < 21 Rightarrow B=1$$ $$ A*BB* (BC)^2 = A*11*(1C)^2 = (EF)^2-1 = (EF-1)(EF+1)$$ which means that either $EF-1$ or $EF+1$ is divisible by $11$ (since $11$ is prime).

This leaves the possibilities for $EF$ as being $23$, $32$, $34$, $43$, $45$, $54$, $56$, $65$, $67$, $76$, $78$, $87$, $89$ and $98$.

We can rule out those which are adjacent to numbers which are divisible by primes greater than $19$ since the left-hand side cannot have such a factor (that is $32$, $45$, $54$, $78$, $87$ and $98$) which leaves $8$ possibilities for $EF$ $rightarrow$ $23$, $34$, $43$, $56$, $65$, $67$, $76$ and $89$.

Furthermore, $frac(EF-1)(EF+1)11$ must be divisible by the square of a number $>11$.

Since $gcd(EF-1, EF+1) leq 2$, this means that either $EF-1$ or $EF+1$ is divisible by a square $>11^2$ (not possible) when $EF$ is even or that either $EF-1$ or $EF+1$ is divisible by an odd square when $EF$ is odd.

This is only true in one case, $EF=89$ and here, we find a solution.

$EF = 89 Rightarrow (EF-1)(EF+1) = 2^4*3^2*5*11 Rightarrow A*(1C)^2 = 2^4*3^2*5$ $$ Rightarrow A=5, ,,,C=2$$

share|improve this answer

edited Jun 26 at 10:41

answered Jun 26 at 10:37

hexominohexomino

55.1k55 gold badges161161 silver badges253253 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Excellent deduction..famous Relationship can be seen from slight rearrangement of terms
  $endgroup$
  – Uvc
  Jun 26 at 10:39
  

  
  
  
  

add a comment | 

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12

$begingroup$

Answer

$$(A,B,C,E,F) = (5,1,2,8,9)$$

Reasoning

$$A * BB geq 22$$ $$ (EF)^2 - B leq 98^2-1 = 9603$$ $$ Rightarrow (BC)^2 leq frac960322 = frac8732 = 436 frac12$$ $$ Rightarrow BC < 21 Rightarrow B=1$$ $$ A*BB* (BC)^2 = A*11*(1C)^2 = (EF)^2-1 = (EF-1)(EF+1)$$ which means that either $EF-1$ or $EF+1$ is divisible by $11$ (since $11$ is prime).

This leaves the possibilities for $EF$ as being $23$, $32$, $34$, $43$, $45$, $54$, $56$, $65$, $67$, $76$, $78$, $87$, $89$ and $98$.

We can rule out those which are adjacent to numbers which are divisible by primes greater than $19$ since the left-hand side cannot have such a factor (that is $32$, $45$, $54$, $78$, $87$ and $98$) which leaves $8$ possibilities for $EF$ $rightarrow$ $23$, $34$, $43$, $56$, $65$, $67$, $76$ and $89$.

Furthermore, $frac(EF-1)(EF+1)11$ must be divisible by the square of a number $>11$.

Since $gcd(EF-1, EF+1) leq 2$, this means that either $EF-1$ or $EF+1$ is divisible by a square $>11^2$ (not possible) when $EF$ is even or that either $EF-1$ or $EF+1$ is divisible by an odd square when $EF$ is odd.

This is only true in one case, $EF=89$ and here, we find a solution.

$EF = 89 Rightarrow (EF-1)(EF+1) = 2^4*3^2*5*11 Rightarrow A*(1C)^2 = 2^4*3^2*5$ $$ Rightarrow A=5, ,,,C=2$$

share|improve this answer

edited Jun 26 at 10:41

answered Jun 26 at 10:37

hexominohexomino

55.1k55 gold badges161161 silver badges253253 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Excellent deduction..famous Relationship can be seen from slight rearrangement of terms
  $endgroup$
  – Uvc
  Jun 26 at 10:39
  

  
  
  
  

add a comment | 

12

$begingroup$

Answer

$$(A,B,C,E,F) = (5,1,2,8,9)$$

Reasoning

$$A * BB geq 22$$ $$ (EF)^2 - B leq 98^2-1 = 9603$$ $$ Rightarrow (BC)^2 leq frac960322 = frac8732 = 436 frac12$$ $$ Rightarrow BC < 21 Rightarrow B=1$$ $$ A*BB* (BC)^2 = A*11*(1C)^2 = (EF)^2-1 = (EF-1)(EF+1)$$ which means that either $EF-1$ or $EF+1$ is divisible by $11$ (since $11$ is prime).

This leaves the possibilities for $EF$ as being $23$, $32$, $34$, $43$, $45$, $54$, $56$, $65$, $67$, $76$, $78$, $87$, $89$ and $98$.

We can rule out those which are adjacent to numbers which are divisible by primes greater than $19$ since the left-hand side cannot have such a factor (that is $32$, $45$, $54$, $78$, $87$ and $98$) which leaves $8$ possibilities for $EF$ $rightarrow$ $23$, $34$, $43$, $56$, $65$, $67$, $76$ and $89$.

Furthermore, $frac(EF-1)(EF+1)11$ must be divisible by the square of a number $>11$.

Since $gcd(EF-1, EF+1) leq 2$, this means that either $EF-1$ or $EF+1$ is divisible by a square $>11^2$ (not possible) when $EF$ is even or that either $EF-1$ or $EF+1$ is divisible by an odd square when $EF$ is odd.

This is only true in one case, $EF=89$ and here, we find a solution.

$EF = 89 Rightarrow (EF-1)(EF+1) = 2^4*3^2*5*11 Rightarrow A*(1C)^2 = 2^4*3^2*5$ $$ Rightarrow A=5, ,,,C=2$$

share|improve this answer

edited Jun 26 at 10:41

answered Jun 26 at 10:37

hexominohexomino

55.1k55 gold badges161161 silver badges253253 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Excellent deduction..famous Relationship can be seen from slight rearrangement of terms
  $endgroup$
  – Uvc
  Jun 26 at 10:39
  

  
  
  
  

add a comment | 

$begingroup$

Answer

$$(A,B,C,E,F) = (5,1,2,8,9)$$

Reasoning

$$A * BB geq 22$$ $$ (EF)^2 - B leq 98^2-1 = 9603$$ $$ Rightarrow (BC)^2 leq frac960322 = frac8732 = 436 frac12$$ $$ Rightarrow BC < 21 Rightarrow B=1$$ $$ A*BB* (BC)^2 = A*11*(1C)^2 = (EF)^2-1 = (EF-1)(EF+1)$$ which means that either $EF-1$ or $EF+1$ is divisible by $11$ (since $11$ is prime).

This leaves the possibilities for $EF$ as being $23$, $32$, $34$, $43$, $45$, $54$, $56$, $65$, $67$, $76$, $78$, $87$, $89$ and $98$.

We can rule out those which are adjacent to numbers which are divisible by primes greater than $19$ since the left-hand side cannot have such a factor (that is $32$, $45$, $54$, $78$, $87$ and $98$) which leaves $8$ possibilities for $EF$ $rightarrow$ $23$, $34$, $43$, $56$, $65$, $67$, $76$ and $89$.

Furthermore, $frac(EF-1)(EF+1)11$ must be divisible by the square of a number $>11$.

Since $gcd(EF-1, EF+1) leq 2$, this means that either $EF-1$ or $EF+1$ is divisible by a square $>11^2$ (not possible) when $EF$ is even or that either $EF-1$ or $EF+1$ is divisible by an odd square when $EF$ is odd.

This is only true in one case, $EF=89$ and here, we find a solution.

$EF = 89 Rightarrow (EF-1)(EF+1) = 2^4*3^2*5*11 Rightarrow A*(1C)^2 = 2^4*3^2*5$ $$ Rightarrow A=5, ,,,C=2$$

share|improve this answer

edited Jun 26 at 10:41

answered Jun 26 at 10:37

hexominohexomino

55.1k55 gold badges161161 silver badges253253 bronze badges

$endgroup$

share|improve this answer

edited Jun 26 at 10:41

answered Jun 26 at 10:37

hexominohexomino

55.1k55 gold badges161161 silver badges253253 bronze badges

answered Jun 26 at 10:37

hexominohexomino

55.1k55 gold badges161161 silver badges253253 bronze badges

answered Jun 26 at 10:37

hexominohexomino

55.1k55 gold badges161161 silver badges253253 bronze badges