Ttdfjt

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We are two immediate neighbors who forged our own powers to form concatenated relationship. Who are we?

The largest Tuesday numberThe largest Friday numberRed, yellow and orange numbersWhat is my four digit car number?Three positive integersThe mysterious self-describing number #2The Special NumbersMadam I m Adam..please don’t get mad..you will no longer be primeHe said yes..she said no..they went back and forth..they agreed toSolve for the Number in the number..square..cube relationship

16

1

$begingroup$

Our concatenated number is $ overlineABAC, $ where $ A, B, C $ are all positive digits (1 - 9).

Our relationship is

$$ overlineABAC = A^A + B^B + A^A + C^C $$

Who are we?

mathematics logical-deduction no-computers number-theory

share|improve this question

edited May 10 at 13:31

Rand al'Thor

72k14238479

asked May 9 at 15:45

UvcUvc

48210

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Having read this question which appeared in the HNQ just today made this puzzle very easy ;-)
  $endgroup$
  – Christoph
  May 9 at 19:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Christoph. What is HNQ?
  $endgroup$
  – Uvc
  May 9 at 19:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Hot Network Questions, a selection of questions from the whole Stack Exchange network which are "featured" to appear in the right-hand sidebar.
  $endgroup$
  – Rand al'Thor
  May 9 at 19:59
  

  
  
  
  

add a comment | 

16

1

$begingroup$

Our concatenated number is $ overlineABAC, $ where $ A, B, C $ are all positive digits (1 - 9).

Our relationship is

$$ overlineABAC = A^A + B^B + A^A + C^C $$

Who are we?

mathematics logical-deduction no-computers number-theory

share|improve this question

edited May 10 at 13:31

Rand al'Thor

72k14238479

asked May 9 at 15:45

UvcUvc

48210

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Having read this question which appeared in the HNQ just today made this puzzle very easy ;-)
  $endgroup$
  – Christoph
  May 9 at 19:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Christoph. What is HNQ?
  $endgroup$
  – Uvc
  May 9 at 19:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Hot Network Questions, a selection of questions from the whole Stack Exchange network which are "featured" to appear in the right-hand sidebar.
  $endgroup$
  – Rand al'Thor
  May 9 at 19:59
  

  
  
  
  

add a comment | 

$begingroup$

Our concatenated number is $ overlineABAC, $ where $ A, B, C $ are all positive digits (1 - 9).

Our relationship is

$$ overlineABAC = A^A + B^B + A^A + C^C $$

Who are we?

mathematics logical-deduction no-computers number-theory

share|improve this question

edited May 10 at 13:31

Rand al'Thor

72k14238479

asked May 9 at 15:45

UvcUvc

48210

$endgroup$

share|improve this question

edited May 10 at 13:31

Rand al'Thor

72k14238479

asked May 9 at 15:45

UvcUvc

48210

share|improve this question

edited May 10 at 13:31

Rand al'Thor

72k14238479

asked May 9 at 15:45

UvcUvc

48210

edited May 10 at 13:31

Rand al'Thor

72k14238479

edited May 10 at 13:31

Rand al'Thor

72k14238479

asked May 9 at 15:45

UvcUvc

48210

asked May 9 at 15:45

UvcUvc

48210

2 Answers
2

active

oldest

votes

21

$begingroup$

Initial bounds

• 
  

  $6^6$ is too big as it has 5 digits.

  
  




• 
  

  $4^4$ is only 256, too small if everything was at most that.

  
  

So at least one of $A,B,C$ must be

$5$.

Narrowing possibilities

• 
  

  $B$ must be even, because $ABAC$ and $C^C$ have matching parity. So $B=2$ or $B=4$.

  
  




• 
  

  If $A=5$, then $A^A+A^A$ is already bigger than $6000$, too big. But $5^5=3125$ is involved somewhere, so $A$ must be at least $3$. So $A=3$ or $A=4$.

  
  

So the only option

to be $5$ is $C$.

Final answer

• 
  

  Trying $A=4$ gives $4B45=3637+B^B$, which is impossible since $2^2$ and $4^4$ are too small to get that high.

  
  




• 
  

  Trying $A=3$ gives $3B35=3179+B^B$, which works with $B=4$.

  
  

So the final answer is

$A=3,B=4,C=5$.

share|improve this answer

edited May 9 at 16:08

answered May 9 at 16:02

Rand al'ThorRand al'Thor

72k14238479

$endgroup$

add a comment | 

20

$begingroup$

Here is the solution

$A=3, B=4, C=5$

$3^3 + 4^4 + 3^3 + 5^5 = 3435$

Reasoning

$2 times 4^4 = 512$ and $6^6=46656$ which is more than four digits so at least one of the digits has to be $5$ and the others have to be less than or equal to $5$.
Then, we have $5^5=3125$ and since double this is more than $6000$, the digit $5$ can appear only once. Furthermore, since $3125 + 2 times 4^4 < 4000$, the digit $3$ must also appear at least once.
Since $5^5 + 3^3 = 3152$, the other unique digit to appear must either be $1$ or $4$ (as this will necessarily be the $2$nd digit). This leaves a small number of possibilities to try.

share|improve this answer

edited May 9 at 15:53

answered May 9 at 15:48

hexominohexomino

49.5k4146235

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Nice, you beat me to it. I think I got a slightly slicker solution by deducing some extra conditions required on $A$ and $B$.
  $endgroup$
  – Rand al'Thor
  May 9 at 16:05
  

  
  
  
  

add a comment | 

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21

$begingroup$

Initial bounds

• 
  

  $6^6$ is too big as it has 5 digits.

  
  




• 
  

  $4^4$ is only 256, too small if everything was at most that.

  
  

So at least one of $A,B,C$ must be

$5$.

Narrowing possibilities

• 
  

  $B$ must be even, because $ABAC$ and $C^C$ have matching parity. So $B=2$ or $B=4$.

  
  




• 
  

  If $A=5$, then $A^A+A^A$ is already bigger than $6000$, too big. But $5^5=3125$ is involved somewhere, so $A$ must be at least $3$. So $A=3$ or $A=4$.

  
  

So the only option

to be $5$ is $C$.

Final answer

• 
  

  Trying $A=4$ gives $4B45=3637+B^B$, which is impossible since $2^2$ and $4^4$ are too small to get that high.

  
  




• 
  

  Trying $A=3$ gives $3B35=3179+B^B$, which works with $B=4$.

  
  

So the final answer is

$A=3,B=4,C=5$.

share|improve this answer

edited May 9 at 16:08

answered May 9 at 16:02

Rand al'ThorRand al'Thor

72k14238479

$endgroup$

add a comment | 

21

$begingroup$

Initial bounds

• 
  

  $6^6$ is too big as it has 5 digits.

  
  




• 
  

  $4^4$ is only 256, too small if everything was at most that.

  
  

So at least one of $A,B,C$ must be

$5$.

Narrowing possibilities

• 
  

  $B$ must be even, because $ABAC$ and $C^C$ have matching parity. So $B=2$ or $B=4$.

  
  




• 
  

  If $A=5$, then $A^A+A^A$ is already bigger than $6000$, too big. But $5^5=3125$ is involved somewhere, so $A$ must be at least $3$. So $A=3$ or $A=4$.

  
  

So the only option

to be $5$ is $C$.

Final answer

• 
  

  Trying $A=4$ gives $4B45=3637+B^B$, which is impossible since $2^2$ and $4^4$ are too small to get that high.

  
  




• 
  

  Trying $A=3$ gives $3B35=3179+B^B$, which works with $B=4$.

  
  

So the final answer is

$A=3,B=4,C=5$.

share|improve this answer

edited May 9 at 16:08

answered May 9 at 16:02

Rand al'ThorRand al'Thor

72k14238479

$endgroup$

add a comment | 

$begingroup$

Initial bounds

• 
  

  $6^6$ is too big as it has 5 digits.

  
  




• 
  

  $4^4$ is only 256, too small if everything was at most that.

  
  

So at least one of $A,B,C$ must be

$5$.

Narrowing possibilities

• 
  

  $B$ must be even, because $ABAC$ and $C^C$ have matching parity. So $B=2$ or $B=4$.

  
  




• 
  

  If $A=5$, then $A^A+A^A$ is already bigger than $6000$, too big. But $5^5=3125$ is involved somewhere, so $A$ must be at least $3$. So $A=3$ or $A=4$.

  
  

So the only option

to be $5$ is $C$.

Final answer

• 
  

  Trying $A=4$ gives $4B45=3637+B^B$, which is impossible since $2^2$ and $4^4$ are too small to get that high.

  
  




• 
  

  Trying $A=3$ gives $3B35=3179+B^B$, which works with $B=4$.

  
  

So the final answer is

$A=3,B=4,C=5$.

share|improve this answer

edited May 9 at 16:08

answered May 9 at 16:02

Rand al'ThorRand al'Thor

72k14238479

$endgroup$

share|improve this answer

edited May 9 at 16:08

answered May 9 at 16:02

Rand al'ThorRand al'Thor

72k14238479

answered May 9 at 16:02

Rand al'ThorRand al'Thor

72k14238479

answered May 9 at 16:02

Rand al'ThorRand al'Thor

72k14238479

20

$begingroup$

Here is the solution

$A=3, B=4, C=5$

$3^3 + 4^4 + 3^3 + 5^5 = 3435$

Reasoning

$2 times 4^4 = 512$ and $6^6=46656$ which is more than four digits so at least one of the digits has to be $5$ and the others have to be less than or equal to $5$.
Then, we have $5^5=3125$ and since double this is more than $6000$, the digit $5$ can appear only once. Furthermore, since $3125 + 2 times 4^4 < 4000$, the digit $3$ must also appear at least once.
Since $5^5 + 3^3 = 3152$, the other unique digit to appear must either be $1$ or $4$ (as this will necessarily be the $2$nd digit). This leaves a small number of possibilities to try.

share|improve this answer

edited May 9 at 15:53

answered May 9 at 15:48

hexominohexomino

49.5k4146235

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Nice, you beat me to it. I think I got a slightly slicker solution by deducing some extra conditions required on $A$ and $B$.
  $endgroup$
  – Rand al'Thor
  May 9 at 16:05
  

  
  
  
  

add a comment | 

20

$begingroup$

Here is the solution

$A=3, B=4, C=5$

$3^3 + 4^4 + 3^3 + 5^5 = 3435$

Reasoning

$2 times 4^4 = 512$ and $6^6=46656$ which is more than four digits so at least one of the digits has to be $5$ and the others have to be less than or equal to $5$.
Then, we have $5^5=3125$ and since double this is more than $6000$, the digit $5$ can appear only once. Furthermore, since $3125 + 2 times 4^4 < 4000$, the digit $3$ must also appear at least once.
Since $5^5 + 3^3 = 3152$, the other unique digit to appear must either be $1$ or $4$ (as this will necessarily be the $2$nd digit). This leaves a small number of possibilities to try.

share|improve this answer

edited May 9 at 15:53

answered May 9 at 15:48

hexominohexomino

49.5k4146235

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Nice, you beat me to it. I think I got a slightly slicker solution by deducing some extra conditions required on $A$ and $B$.
  $endgroup$
  – Rand al'Thor
  May 9 at 16:05
  

  
  
  
  

add a comment | 

$begingroup$

Here is the solution

$A=3, B=4, C=5$

$3^3 + 4^4 + 3^3 + 5^5 = 3435$

Reasoning

$2 times 4^4 = 512$ and $6^6=46656$ which is more than four digits so at least one of the digits has to be $5$ and the others have to be less than or equal to $5$.
Then, we have $5^5=3125$ and since double this is more than $6000$, the digit $5$ can appear only once. Furthermore, since $3125 + 2 times 4^4 < 4000$, the digit $3$ must also appear at least once.
Since $5^5 + 3^3 = 3152$, the other unique digit to appear must either be $1$ or $4$ (as this will necessarily be the $2$nd digit). This leaves a small number of possibilities to try.

share|improve this answer

edited May 9 at 15:53

answered May 9 at 15:48

hexominohexomino

49.5k4146235

$endgroup$

share|improve this answer

edited May 9 at 15:53

answered May 9 at 15:48

hexominohexomino

49.5k4146235

answered May 9 at 15:48

hexominohexomino

49.5k4146235

answered May 9 at 15:48

hexominohexomino

49.5k4146235