Welcome to our community

Be a part of something great, join today!

Solve in positive integers

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
Solve in positive integers for

$577(bcd+b+c)=520(abcd+ab+ac+cd+1)$
 

mente oscura

Well-known member
Nov 29, 2013
172
Solve in positive integers for

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

Merry christmas. (Party)

[tex]577(bcd+b+c)=520a(bcd+b+c)+520(cd+1)[/tex]

[tex](577-520a)(bcd+b+c)=+520(cd+1)[/tex]

First conclusion: [tex]\ a=1[/tex]

[tex]57(bcd+b+c)=+520(cd+1)[/tex]

[tex]57b(cd+1)+57c=+520(cd+1)[/tex]

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

[tex]57 \ and \ 520 \ coprime[/tex]

[tex]57|(cd+1)[/tex]

A bit of brute force. :eek:

[tex]a=1[/tex]
[tex]b=9[/tex]
[tex]c=7[/tex]
[tex]d=8[/tex]


Regards.
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
Hello. (Sun)

Merry christmas. (Party)

[tex]577(bcd+b+c)=520a(bcd+b+c)+520(cd+1)[/tex]

[tex](577-520a)(bcd+b+c)=+520(cd+1)[/tex]

First conclusion: [tex]\ a=1[/tex]

[tex]57(bcd+b+c)=+520(cd+1)[/tex]

[tex]57b(cd+1)+57c=+520(cd+1)[/tex]

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

[tex]57 \ and \ 520 \ coprime[/tex]

[tex]57|(cd+1)[/tex]

A bit of brute force. :eek:

[tex]a=1[/tex]
[tex]b=9[/tex]
[tex]c=7[/tex]
[tex]d=8[/tex]


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

so b = 9
 

mente oscura

Well-known member
Nov 29, 2013
172
57c=(520−57b)(cd+1)
=> 520 - 57b < 57
or b >= 9

so b = 9
[tex]57c=7(cd+1)[/tex]

[tex]c(57-7d)=7 \ \rightarrow{d=8} \ \rightarrow{c=7}[/tex]


Regards.
 
  • Thread starter
  • Admin
  • #5

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
Hello. (Sun)

Merry christmas. (Party)

[tex]577(bcd+b+c)=520a(bcd+b+c)+520(cd+1)[/tex]

[tex](577-520a)(bcd+b+c)=+520(cd+1)[/tex]

First conclusion: [tex]\ a=1[/tex]

[tex]57(bcd+b+c)=+520(cd+1)[/tex]

[tex]57b(cd+1)+57c=+520(cd+1)[/tex]

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

[tex]57 \ and \ 520 \ coprime[/tex]

[tex]57|(cd+1)[/tex]

A bit of brute force. :eek:

[tex]a=1[/tex]
[tex]b=9[/tex]
[tex]c=7[/tex]
[tex]d=8[/tex]


Regards.
[tex]57c=7(cd+1)[/tex]

[tex]c(57-7d)=7 \ \rightarrow{d=8} \ \rightarrow{c=7}[/tex]


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!