The sum of two natural numbers when added to their LCM gives a total of 143. How many such pairs of numbers exist?

1) 9
2) 8
3) 10
4) 7
5) 6

• Let the numbers be mx and my where x and y are co prime.
  Then, their LCM = mxy
  
  Given: mx + my + mxy = 143
  
  => m(x+y+xy) = 143
  
  Add m on both sides:
  => m(x+y+xy+1) = 143+m
  => m(x+1)(y+1) = 143 +m
  => (x+1)(y+1) = 143/m + 1
  
  So, m as a factor of 143 can be 1,11,13, 143.
  
  Case 1: m = 1, then (x+1)(y+1) = 144
  
  144
  = 2 * 72 or (x, y) = (1, 71)
  = 3 * 48 or (x, y) = (2, 47)
  = 4 * 36 or (x, y) = (3, 35)
  = 6 * 24 or (x, y) = (5, 23)
  = 8 * 18 or (x, y) = (7,17)
  = 9 * 16 or (x, y) = (8, 15)
  = 12 * 12 or (x, y) = (11, 11)
  
  
  Case 2: m = 11, then (x+1)(y+1) = 14
  
  14
  = 2 * 7 or (x, y) = (1, 6) or the numbers are: (11, 66)
  
  
  Case 3: m = 13, then (x+1)(y+1) = 12
  
  12
  = 2 * 6 or (x, y) = (1, 5) or the numbers are: (13, 65)
  = 3 * 4 or (x, y) = (2, 3) or the numbers are: (26, 39)
  
  Case 4: m = 143, then (x+1)(y+1) = 2
  No solution.
  
  
  Hence, overall there are 9 solutions -
  (1, 71), (2,47), (3, 35), (5, 23), (7, 17), (8, 15), (11, 66), (13, 65), (26, 39)
  
  Ans: Option (A)
• 6 years agoHelpfull: Yes(0) No(0)
• let the numbers be mx and my with x and y co prime
  
  so LCM = mxy
  
  so mx + my + mxy = 143
  
  or m(x+y+xy) = 143
  
  add m on both sides
  
  m(x+y+xy+1) = 143+m
  
  m(x+1)(y+1) = 143 +m or (x+1)(y+1) = 143/m + 1
  
  so m is factor or 143 can be 1,11,13, 143
  
  m = 1=> (x+1)(y+1) = 144
  144 = 2 * 72( x=1 y = 71)
  = 3 * 48 (2,47)
  = 4 * 36(3,35)
  = 6 * 24( 5,23)
  = 8 * 18 (7,17)
  = 9 * 16(8,15)
  = 12 * 12 (11,11)
  all except (11,11) are coprimes so solution
  
  m = 11 => (x+1)(y+1) = 14 , x + 1 = 2, x = 1 and y+1 = 7 y = 6 so 11 and 66
  m = 13 => (x+1)(y+1) = 12 (x=1, y = 5), (x=2, y = 3) so 13, 65 or 26,39
  m = 143 => (x+1)(y+1) = 2 so no solution
  
  so 9 solutions (1,71),(5,23), (8,15), (7,17),(3,35), (2,47) (11,66), (13,65), (26,39)
• 2 years agoHelpfull: Yes(0) No(0)