# Solve in positive integers

#### anemone

##### MHB POTW Director
Staff member
Solve in positive integers for

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

#### mente oscura

##### Well-known member
Solve in positive integers for

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

Merry christmas.

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

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

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.

##### Well-known member
Hello.

Merry christmas.

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

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

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

so b = 9

#### mente oscura

##### Well-known member
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.

#### anemone

##### MHB POTW Director
Staff member
Hello.

Merry christmas.

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

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

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.
$$57c=7(cd+1)$$

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

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!

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

so b = 9
Yes, that's correct, kaliprasad and thanks for participating!