when m+n is divided by 12 the reminder is 8 and m-n is divided by 12 the reminder is 6.what is the reminder when mn is divided by 6??

• according to the given condition
  m+n = 12x+8
  m-n = 12y+6
  on solving we get m = 6x+6y+7, n = 6x-6y+1
  so mn = (6x+6y+7)(6x-6y+1) = 36(x^2-y^2)+6x+6y+42x-42y+7
  this expression is equal to 6z+1 so on dividing it by 6 we will get 1 as reminder
• 12 years agoHelpfull: Yes(34) No(4)
• Remainder is always 1
  This answer can be proved with multiple values... 2 of them are :-
  1. m=7 n=1
  2. m=19 n=1
  
  For m=7,n=1
  m+n = 8 divide by 12 gives remainder 8
  m-n = 6 divide by 12 gives remainder 6
  so, m*n = 7 divide by 6 gives remainder 1
  
  
  For m=19,n=1
  m+n = 20 divide by 12 gives remainder 8
  m-n = 18 divide by 12 gives remainder 6
  so, m*n = 19 divide by 6 gives remainder 1
• 12 years agoHelpfull: Yes(30) No(0)
• answer is 1
  here m is 19 and n is 1.
  m+n = 20 leaves 8 when divided by 12.
  m-n = 18 leaves 6 when divided by 12.
  mn = 19 leaves 1 when divided by 6.
  
• 12 years agoHelpfull: Yes(0) No(0)
• 7 and 1
  
  (7+1)%12=8
  (7-1)%12=6
• 12 years agoHelpfull: Yes(0) No(1)
• this can be done with assuming two no.
  49+7=56 give remainder 8
  49-7=42 give remainder 6
  49*7/6 remainder =1
• 12 years agoHelpfull: Yes(0) No(0)
• Since the remainder when m+n is divided by 12 is 8, m+n=12p+8;
  and since the remainder when m–n is divided by 12 is 6, m–n=12q+6.
  
  Here, p and q are integers. Adding the two equations yields 2m=12p+12q+14.
  Solving for m yields m=6p+6q+7=6(p+q+1)+1=6r+1,
  where r is a positive integer equalling p+q+1.
  
  Now, let’s subtract the equations m+n=12p+8 and m–n=12q+6.
  This yields 2n=(12p+8)–(12q+6)=12(p–q)+2.
  
  Solving for n yields n=6(p–q)+1=6t+1,
  where t is an integer equalling p–q
  
  Hence, we have mn=(6r+1)(6t+1)=36rt+6r+6t+1=6(6rt+r+t)+1 by factoring out 6
  
  Hence, the remainder is 1
• 11 years agoHelpfull: Yes(0) No(0)