Two numbers when divided by a certain divisor leave remainders of 431 and 379 respectively. When the sum of these two numbers is divided by the same divisor, the remainder is 211. What is the divisor?

• 599?
  
  x = 431 mod d
  y = 379 mod d
  
  therefore,
  (x+y) = (431 + 379) mod d
  
  as is given remainder from sum is 211
  therefore, d divides (431 + 379) - 211 = 599
  
  now 599 is prime number so d = 599
• 7 years agoHelpfull: Yes(5) No(5)
• X=431mod d
  Y=379mod d
  X+y=(431+379)mod d
  given remainder from sum is 211
  therefore, d divides (431 + 379) - 211 = 599
  Divisor = (431+379)-211=599
  Divisor =599
• 6 years agoHelpfull: Yes(1) No(0)
• 599 is the Answer..
  
  Let A be the divisor, X and Y be the two numbers. ( A is an integer)
  We have: X= Ax + 431
  Y= Ay+ 379
  X+Y= Az + 211= A(x+y) + 810 ( x, y and z and integers)
  ---> A ( z-x-y)= 810-211= 599
• 6 years agoHelpfull: Yes(1) No(0)
• Let A be the divisor,
  X and Y be the two numbers.
  X = Ax + 431
  Y = Ay + 379
  X + Y = Az + 211 = A(x + y) + 810
  or A (z – x – y) = 810 – 211 = 599
• 5 years agoHelpfull: Yes(0) No(0)