sum of two numbers is 483 and their hcf is 23.find themaximum possiblenumber of pairs that satisfy the condition?

• 6 pairs
  if numbers are 23x and 23y where x and y are relatively prime then
  23*(x+y)=483
  x+y= 21
  so possible pairs for (x,y) are
  1,20
  2,19,
  4,17
  5,16
  8,13
  10,11
  
  you can interchange values of x and y .
• 10 years agoHelpfull: Yes(8) No(1)
• 6 pairs..cause 23(x+y)=(23*21)..so x+y=21.there are only 6 ways for numbers to be not a multiple of some number....(1,20),(2,19),(4,17),(5,16),(8,13),(10,11)
• 10 years agoHelpfull: Yes(3) No(1)
• given that x+y=483 and hcf=23.Then obviously
  23x+23y=483
  23(x+y)=483
  x+y=483/23
  x+y=21
  the values of x and y which satisfies 21 are
  (1,20),(2,19),(3,18)(4,17),(5,16),(6,15).so no. of pairs are 6.
• 10 years agoHelpfull: Yes(0) No(2)
• 6 pairs, (23,21),(21,23),(1,483),(3,161)(483,1),(161,3)
• 10 years agoHelpfull: Yes(0) No(0)
• it will be 483/23 = 21
  now 21/2 = 10
  ans is 10
• 10 years agoHelpfull: Yes(0) No(0)
• 10 pairs
  if numbers are 23n and 23m and 23n+23m=483
  n+m=21
  the possible pairs are (1,20)(2,19)(3,18)(4,17)(5,16)(6,15)(7,14)(8,13)(9,12)(10,11)
• 10 years agoHelpfull: Yes(0) No(0)