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: $$\ a=1$$

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

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

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

$$57 \ and \ 520 \ coprime$$

$$57|(cd+1)$$

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!