Etydhv

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

What is my four digit car number?prove that 2016 is a self-composable numberOh dang, I have to change my whole answer!The mysterious self-describing number #2The Special NumbersPowers of two my children beMadam 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 relationshipLong digital sequence. 16xxxxxxxxxxxxx61

6

$begingroup$

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

share|improve this question

edited 2 hours ago

PiIsNot3

3,881847

asked 4 hours ago

UvcUvc

4457

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Having read this question which appeared in the HNQ just today made this puzzle very easy ;-)
  $endgroup$
  – Christoph
  57 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Christoph. What is HNQ?
  $endgroup$
  – Uvc
  46 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  10 mins ago
  

  
  
  
  

add a comment | 

6

$begingroup$

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

share|improve this question

edited 2 hours ago

PiIsNot3

3,881847

asked 4 hours ago

UvcUvc

4457

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Having read this question which appeared in the HNQ just today made this puzzle very easy ;-)
  $endgroup$
  – Christoph
  57 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Christoph. What is HNQ?
  $endgroup$
  – Uvc
  46 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  10 mins ago
  

  
  
  
  

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

share|improve this question

edited 2 hours ago

PiIsNot3

3,881847

asked 4 hours ago

UvcUvc

4457

$endgroup$

share|improve this question

edited 2 hours ago

PiIsNot3

3,881847

asked 4 hours ago

UvcUvc

4457

share|improve this question

edited 2 hours ago

PiIsNot3

3,881847

asked 4 hours ago

UvcUvc

4457

edited 2 hours ago

PiIsNot3

3,881847

edited 2 hours ago

PiIsNot3

3,881847

asked 4 hours ago

UvcUvc

4457

asked 4 hours ago

UvcUvc

4457

2 Answers
2

active

oldest

votes

7

$begingroup$

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 4 hours ago

answered 4 hours ago

hexominohexomino

49.2k4146231

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  4 hours ago
  

  
  
  
  

add a comment | 

7

$begingroup$

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 4 hours ago

answered 4 hours ago

Rand al'ThorRand al'Thor

71.6k14238477

$endgroup$

add a comment | 

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7

$begingroup$

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 4 hours ago

answered 4 hours ago

hexominohexomino

49.2k4146231

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  4 hours ago
  

  
  
  
  

add a comment | 

7

$begingroup$

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 4 hours ago

answered 4 hours ago

hexominohexomino

49.2k4146231

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  4 hours ago
  

  
  
  
  

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 4 hours ago

answered 4 hours ago

hexominohexomino

49.2k4146231

$endgroup$

share|improve this answer

edited 4 hours ago

answered 4 hours ago

hexominohexomino

49.2k4146231

answered 4 hours ago

hexominohexomino

49.2k4146231

answered 4 hours ago

hexominohexomino

49.2k4146231

7

$begingroup$

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 4 hours ago

answered 4 hours ago

Rand al'ThorRand al'Thor

71.6k14238477

$endgroup$

add a comment | 

7

$begingroup$

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 4 hours ago

answered 4 hours ago

Rand al'ThorRand al'Thor

71.6k14238477

$endgroup$

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 4 hours ago

answered 4 hours ago

Rand al'ThorRand al'Thor

71.6k14238477

$endgroup$

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Rand al'ThorRand al'Thor

71.6k14238477

answered 4 hours ago

Rand al'ThorRand al'Thor

71.6k14238477

answered 4 hours ago

Rand al'ThorRand al'Thor

71.6k14238477