MHB What is the solution to this equation in positive integers?

[anemone]

Solve in positive integers for

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

 

[mente oscura]

Solve in positive integers for

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

Hello. (Sun)

Merry christmas. (Party)

Spoiler

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.

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

Regards.

 

[kaliprasad]

Hello. (Sun)

Merry christmas. (Party)

Spoiler

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.

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

Regards.

Spoiler

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

so b = 9

 

[mente oscura]

Spoiler

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

so b = 9

Spoiler

57c=7(cd+1)

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

Regards.

 

[anemone]

Hello. (Sun)

Merry christmas. (Party)

Spoiler

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.

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

Regards.

Spoiler

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!

Spoiler

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

so b = 9

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