The sum of two numbers is 462 and their HCF is 22. What is the maximum possible number of pairs that satisfy these conditions?

1) 1
2) 3
3) 2
4) 6
5) 8

• 22(x+y) = 462
  x+y will be 21
  the pairs which will satisfy 22 as hcf are (1,20) (2,19) (4,17) (5,16) (8,13) (10,11)
• 11 years agoHelpfull: Yes(6) No(2)
• if HCF is 22 numbers will be 22x and 22y
  22x+22y=462
  x+y=21
  therefore the set of co-primes will be (1,20)(2,19)(4,17)(5,16)(8,13)(10,11)
  therefore numbers will be(22,440)(44,418)(88,374)(110,352)(176,286)(220,242)
  hence answer is 6.
• 11 years agoHelpfull: Yes(1) No(0)
• First, let the numbers be 22a and 22b.
  
  Therefore, 22(a+b) = 462.
  
  or, a+b = 21. Therefore, the possible pairs are: 1+21, 4+17, 5+16, 8+13, 10+11, i.e. 5.
  
  Remember, a and b are prime to each other, i.e. they do not have any common factor between them.
  You cannot choose 3+18 as one of the pairs as 3 and 18 have 3 as a common factor. Similarly you cannot choose 6+15 as a possible pair. So only 5 pairs are possible.
  
  If u multiply each of those 5 pairs by 22, their HCF will come as 22 and their sum will be 462.
• 8 years agoHelpfull: Yes(1) No(0)