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2

$begingroup$

Given:

A, B, C are 3 distinct primes.

Find the smallest composite number Y that satisfies the relation:

$ Y = A ^ C + B ^ B + C ^ A $

mathematics logical-deduction no-computers

share|improve this question

edited May 16 at 19:37

LeppyR64

11.4k23859

asked May 16 at 17:51

UvcUvc

88414

$endgroup$

add a comment | 

2

$begingroup$

Given:

A, B, C are 3 distinct primes.

Find the smallest composite number Y that satisfies the relation:

$ Y = A ^ C + B ^ B + C ^ A $

mathematics logical-deduction no-computers

share|improve this question

edited May 16 at 19:37

LeppyR64

11.4k23859

asked May 16 at 17:51

UvcUvc

88414

$endgroup$

add a comment | 

$begingroup$

Given:

A, B, C are 3 distinct primes.

Find the smallest composite number Y that satisfies the relation:

$ Y = A ^ C + B ^ B + C ^ A $

mathematics logical-deduction no-computers

share|improve this question

edited May 16 at 19:37

LeppyR64

11.4k23859

asked May 16 at 17:51

UvcUvc

88414

$endgroup$

share|improve this question

edited May 16 at 19:37

LeppyR64

11.4k23859

asked May 16 at 17:51

UvcUvc

88414

share|improve this question

edited May 16 at 19:37

LeppyR64

11.4k23859

asked May 16 at 17:51

UvcUvc

88414

edited May 16 at 19:37

LeppyR64

11.4k23859

edited May 16 at 19:37

LeppyR64

11.4k23859

asked May 16 at 17:51

UvcUvc

88414

asked May 16 at 17:51

UvcUvc

88414

2 Answers
2

active

oldest

votes

8

$begingroup$

The goal is to minimize, so you'll need to obviously use the primes

$2, 3, 5$

Then

The $5$ can't be in the middle as $5^5$ is large, so it must be on either side. $(2^5 + 3^3) < (3^5 + 2^2)$, so your equation is $2^5 + 3^3 + 5^2$.

This means

$(A, B, C) = (2, 3, 5)$ and $Y = 84$

share|improve this answer

answered May 16 at 17:59

AranlydeAranlyde

907213

$endgroup$

add a comment | 

4

$begingroup$

I totally agree with the answer given by @Aranlyde.

But you don't explicitly specify that $Anot=Bnot=C$.

So, I would do something like $A=B=C=2$.

Then, the result would be:

$Y=A^C+B^B+C^A=2^2+2^2+2^2=4+4+4=12$


In any way, this is the minimum.

share|improve this answer

answered May 16 at 18:20

ArdorisArdoris

3015

New contributor

Ardoris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Three prime pals refer to three different..i will edit to add distinct..thx
  $endgroup$
  – Uvc
  May 16 at 18:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Okay, well. Then my answer is incorrect, of course. :)
  $endgroup$
  – Ardoris
  May 16 at 18:32
  

  
  
  
  

add a comment | 

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8

$begingroup$

The goal is to minimize, so you'll need to obviously use the primes

$2, 3, 5$

Then

The $5$ can't be in the middle as $5^5$ is large, so it must be on either side. $(2^5 + 3^3) < (3^5 + 2^2)$, so your equation is $2^5 + 3^3 + 5^2$.

This means

$(A, B, C) = (2, 3, 5)$ and $Y = 84$

share|improve this answer

answered May 16 at 17:59

AranlydeAranlyde

907213

$endgroup$

add a comment | 

8

$begingroup$

The goal is to minimize, so you'll need to obviously use the primes

$2, 3, 5$

Then

The $5$ can't be in the middle as $5^5$ is large, so it must be on either side. $(2^5 + 3^3) < (3^5 + 2^2)$, so your equation is $2^5 + 3^3 + 5^2$.

This means

$(A, B, C) = (2, 3, 5)$ and $Y = 84$

share|improve this answer

answered May 16 at 17:59

AranlydeAranlyde

907213

$endgroup$

add a comment | 

$begingroup$

The goal is to minimize, so you'll need to obviously use the primes

$2, 3, 5$

Then

The $5$ can't be in the middle as $5^5$ is large, so it must be on either side. $(2^5 + 3^3) < (3^5 + 2^2)$, so your equation is $2^5 + 3^3 + 5^2$.

This means

$(A, B, C) = (2, 3, 5)$ and $Y = 84$

share|improve this answer

answered May 16 at 17:59

AranlydeAranlyde

907213

$endgroup$

share|improve this answer

answered May 16 at 17:59

AranlydeAranlyde

907213

answered May 16 at 17:59

AranlydeAranlyde

907213

answered May 16 at 17:59

AranlydeAranlyde

907213

4

$begingroup$

I totally agree with the answer given by @Aranlyde.

But you don't explicitly specify that $Anot=Bnot=C$.

So, I would do something like $A=B=C=2$.

Then, the result would be:

$Y=A^C+B^B+C^A=2^2+2^2+2^2=4+4+4=12$


In any way, this is the minimum.

share|improve this answer

answered May 16 at 18:20

ArdorisArdoris

3015

New contributor

Ardoris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Three prime pals refer to three different..i will edit to add distinct..thx
  $endgroup$
  – Uvc
  May 16 at 18:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Okay, well. Then my answer is incorrect, of course. :)
  $endgroup$
  – Ardoris
  May 16 at 18:32
  

  
  
  
  

add a comment | 

4

$begingroup$

I totally agree with the answer given by @Aranlyde.

But you don't explicitly specify that $Anot=Bnot=C$.

So, I would do something like $A=B=C=2$.

Then, the result would be:

$Y=A^C+B^B+C^A=2^2+2^2+2^2=4+4+4=12$


In any way, this is the minimum.

share|improve this answer

answered May 16 at 18:20

ArdorisArdoris

3015

New contributor

Ardoris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Three prime pals refer to three different..i will edit to add distinct..thx
  $endgroup$
  – Uvc
  May 16 at 18:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Okay, well. Then my answer is incorrect, of course. :)
  $endgroup$
  – Ardoris
  May 16 at 18:32
  

  
  
  
  

add a comment | 

$begingroup$

I totally agree with the answer given by @Aranlyde.

But you don't explicitly specify that $Anot=Bnot=C$.

So, I would do something like $A=B=C=2$.

Then, the result would be:

$Y=A^C+B^B+C^A=2^2+2^2+2^2=4+4+4=12$


In any way, this is the minimum.

share|improve this answer

answered May 16 at 18:20

ArdorisArdoris

3015

New contributor

Ardoris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this answer

answered May 16 at 18:20

ArdorisArdoris

3015

New contributor

Ardoris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered May 16 at 18:20

ArdorisArdoris

3015

New contributor

Ardoris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered May 16 at 18:20

ArdorisArdoris

3015