What is the solution to this equation in positive integers?

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Discussion Overview

The discussion revolves around finding solutions in positive integers for the equation $577(bcd+b+c)=520(abcd+ab+ac+cd+1)$. The scope includes mathematical reasoning and problem-solving related to integer equations.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • Post 1 presents the equation to be solved in positive integers.
  • Subsequent posts include greetings and expressions of goodwill, with minimal engagement on the mathematical problem.
  • Post 5 acknowledges a participant's solution as correct but does not elaborate on the details of the solution or the reasoning behind it.
  • There is a mention of agreement on the correctness of a solution, but no specific details or methods are discussed.

Areas of Agreement / Disagreement

While there is acknowledgment of a correct solution by some participants, the discussion lacks detailed exploration of the methods used or any competing views, leaving the overall resolution of the mathematical problem unclear.

Contextual Notes

The discussion does not provide specific assumptions or mathematical steps that lead to the proposed solutions, which may limit understanding of the problem-solving process.

Who May Find This Useful

Readers interested in integer equations and mathematical problem-solving may find the discussion relevant.

anemone
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Solve in positive integers for

$577(bcd+b+c)=520(abcd+ab+ac+cd+1)$
 
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anemone said:
Solve in positive integers for

$577(bcd+b+c)=520(abcd+ab+ac+cd+1)$

Hello. (Sun)

Merry christmas. (Party)

577(bcd+b+c)=520a(bcd+b+c)+520(cd+1)

(577-520a)(bcd+b+c)=+520(cd+1)

First conclusion: \ a=1

57(bcd+b+c)=+520(cd+1)

57b(cd+1)+57c=+520(cd+1)

57c=(520-57b)(cd+1) \ \rightarrow{b<10}

57 \ and \ 520 \ coprime

57|(cd+1)

A bit of brute force. :o

a=1
b=9
c=7
d=8

Regards.
 
mente oscura said:
Hello. (Sun)

Merry christmas. (Party)

577(bcd+b+c)=520a(bcd+b+c)+520(cd+1)

(577-520a)(bcd+b+c)=+520(cd+1)

First conclusion: \ a=1

57(bcd+b+c)=+520(cd+1)

57b(cd+1)+57c=+520(cd+1)

57c=(520-57b)(cd+1) \ \rightarrow{b<10}

57 \ and \ 520 \ coprime

57|(cd+1)

A bit of brute force. :o

a=1
b=9
c=7
d=8

Regards.

57c=(520−57b)(cd+1)
=> 520 - 57b < 57
or b >= 9

so b = 9
 
kaliprasad said:
57c=(520−57b)(cd+1)
=> 520 - 57b < 57
or b >= 9

so b = 9

57c=7(cd+1)

c(57-7d)=7 \ \rightarrow{d=8} \ \rightarrow{c=7}

Regards.
 
mente oscura said:
Hello. (Sun)

Merry christmas. (Party)

577(bcd+b+c)=520a(bcd+b+c)+520(cd+1)

(577-520a)(bcd+b+c)=+520(cd+1)

First conclusion: \ a=1

57(bcd+b+c)=+520(cd+1)

57b(cd+1)+57c=+520(cd+1)

57c=(520-57b)(cd+1) \ \rightarrow{b&lt;10}

57 \ and \ 520 \ coprime

57|(cd+1)

A bit of brute force. :o

a=1
b=9
c=7
d=8

Regards.

mente oscura said:
57c=7(cd+1)

c(57-7d)=7 \ \rightarrow{d=8} \ \rightarrow{c=7}

Regards.

Bravo, mente oscura! Your answer is correct and your solution is pretty much the same as the one that I have and you're simply awesome!

Merry Christmas to you too!

kaliprasad said:
57c=(520−57b)(cd+1)
=> 520 - 57b < 57
or b >= 9

so b = 9

Yes, that's correct, kaliprasad and thanks for participating!
 

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