(10!+111)/143 what is the remainder

• 143 => 11*13
  
  10! mod 11 + 111 mod 11 = R1
  R[(p-1)!/p] = p-1 [Provided p is a prime no.]... So R[10!/11] = 10...
  10 + 1 mos 11 => 11 mod 11 = 0
  R1 = 0
  
  10! mod 13 + 111 mod 13 = R2
  6 + 7 mod 13
  13 mod 13 = 0
  R2 = 0
  
  Rem = 11x+R1 = 13y+R2
  Rem = 11x+0 = 13y+0
  x = 13, y = 11
  143 mod 143 = 0
  
  Ans : 0
  
• 10 years agoHelpfull: Yes(6) No(0)
• 10!=10*9*8*7*6*5*4*3*2*1
  Now,when (a+b) is devided by n then remainder will be (a/n)+(b/n)
  & when (a*b) is devided by n then remainder will be (a/n)*(b/n)
  So,111/143 will give remainder 111.
  Now,10!=(3*5*10)*(1*2*4*6*7)*(8*9)
  A * B * C
  when A is devided by 143 will give remainder 7
  when B is devided by 143 will give remainder 50
  when C is devided by 143 will give remainder 72
  A*B=350 when devided by 143 gives 64 as remainder.
  64*C=64*72=16*4*9*8=(144)*(32) when devided by 143 gives remainder 1*32=32
  So,finally 32+111=143 when devided by 143 gives remainer 0
  
  
• 10 years agoHelpfull: Yes(0) No(0)