If m+n gives remainder 8 & m-n gives remainder 6 when divided by 12 , what is remainder when mn divided by 6??

• 
  12a+8=m+n;12b+6=m-n;therefore mn=[6(a+b+1)+1]*[(6(a-b)+1];so the remainder is 1 when divided by 6
• 11 years agoHelpfull: Yes(35) No(3)
• m+n=12x+8
  m-n=12y=6
  m=12(x+y)+14=6(2x+2y)+14
  n=12(x-y)+2=6(2x-2y)+2
  mn/6 produces a remainder of 4i.e(14*2)/6
• 11 years agoHelpfull: Yes(18) No(30)
• m+n=12x+8
  m-n=12y+6
  
  solve for m & n
  m=6(x+y)+7
  n=6(x-y)+1
  
  mn/6=1
• 11 years agoHelpfull: Yes(15) No(0)
• it is given that when sum of digit is divided by 12 gives 8 as a remainder
  and their difference divided by 12 gives 6
  
  therefore we have to suppose such numbers whose sum is 20 and difference is 6 by hit n trial we found (13,7) is such pair
  m=13 and n=7
  
  on putting m=13 and n=7 in the given condition we get ~
  
  m*n/6=13*7/6=91/6 yields 1 as a remainder
  
• 11 years agoHelpfull: Yes(11) No(0)
• The ans is=2
  (m+n)*12+8
  (12m+12n)+8
  Let m=5, n=4
  (60+48)+8=116
  116%12=8 (it also satisfy (m-n))
  so m*n=5*4=20
  =20%6=2
• 11 years agoHelpfull: Yes(1) No(16)
• m+n=12x+8.....(1)
  m-n=12y+6.....(2)
  2m=12(x+y)+14 since adding 2 eq's
  m=6(x+y)+7
  2n=12(x-y)+2 since subtracting 2 eq's
  n=6(x-y)+1
  ==>mn=(6(x+y)+7)*(6(x-y)+1)
  mn%6=7*1=7
  ans=7
  
• 11 years agoHelpfull: Yes(1) No(7)
• ans is 1
  m+n=12a+8
  m-n=12b=6
  => m=6(a+b)+7=6k+1
  & n=6(a-b)+1
  therefore, m*n/6 will give remainder as 1
• 11 years agoHelpfull: Yes(1) No(2)
• R(m+n+m-n)/.....=8+6=14/2====7
  M+N-(M-N)=1
  MN=1*7=7
• 11 years agoHelpfull: Yes(0) No(2)
• 4 is the right answer ..
  Let us take m=12 and n=8 which satisfies the above condition ...
  So m*n=96
  96/6 will gives you the remainder 4 ..
  100% solved now ..
  
• 11 years agoHelpfull: Yes(0) No(5)
• here the remainder will be 7 and when it is divided by 6, the remainder will be 1.
• 11 years agoHelpfull: Yes(0) No(0)
• m+n=12a+8
  m-n=12b+6
  2m=12(a+b)+14
  m=6(a+b)+7.......(i)
  2n=12(a-b)+2
  n=6(a-b)+1........(ii)
  mn=[12(a+b)+7][12(a-b)+1]=36(a^2-b^2)+6(a+b)+42(a-b)+7
  mn/6=7/6=rem=1(ans)
• 10 years agoHelpfull: Yes(0) No(0)
• m+n=12x+8
  m-n=12y+6 so solving above two equation we get
  ⇒m=6(x+y)+7 and
  ⇒n=6(x-y)+7
  so mn=[6(x+y)+7][6(x-y)+7]
  ⇒36x^2-36xy+42x+36xy-36y^2+42x+42y-42y+49
  ⇒36x^2-36y^2+84x+49
  when we divide the above equation wrt 6 then remainder is 1.
  
  
• 9 years agoHelpfull: Yes(0) No(0)
• m+n=8
  m-n=6
  a=m+n/2=8+6/2=14/2=7
  b=m-n/2=8-6/2=2/2=1
  therefore, mn = 7-1/6=6/6=1
• 9 years agoHelpfull: Yes(0) No(0)
• mn/6=((m+n)^2/4-(m-n)^2/4)/6 [expanding mn in terms of m+n and m-n];
  replace the value of m+n and m-n with the remainder values [as number divided by 6 will give same remainder as number divided by 12];
  mn/6 = (8^2/4-6^2/4)/6
  i.e, 7/6;
  therefor, remainder of mn/6 i.e, mn%6=1
• 8 years agoHelpfull: Yes(0) No(0)