the least number which when divided by 5,6,7 and 8 leaves a remainder 3,but when divided by 9 leaves no remainder is:

• devided by 5 6 7 8 it leaves 3
  
  so it is 3 more than multiples ot LCM of 5 6 7 8
  
  now LCM of 5 6 7 8
  factorisee
  5 = 5
  6 = 2*3
  7 = 7
  8 = 2* 2* 2 = 2^3
  LCM = 5*2^3*3*7 = 840
  now it has to be of the form 840n + 3
  
  this is divisible by 9
  
  840n + 3 = 9m
  
  3 = 9m - 840n
  
  now using extended euler algorithm
  
  we need to find GCD of 840 and 9
  
  840 = 9*93 + 3
  3= 840*1 - 9*93
  so n = -1 or 2 as it is mod 3
  
  so 840*n +3 = 1683 divisible by 9
  
  hence ans is 1683
  
• 11 years agoHelpfull: Yes(1) No(0)
• 1683
  
  5*6*7*8 + 3 = 1683
• 11 years agoHelpfull: Yes(0) No(1)
• the least number which when divided by 5,6,7&8 leaves a remainder 3 is = (LCM of 5, 6, 7 and 8 + 3)
  LCM of 6 and 8 is 24
  LCM of 5, 7, 24 is = 5 × 7 × 24 = 840
  so the least number = 840 + 3 = 843
  sum of digits 8 + 4 + 3 = 15 not divisible by 9
  Next number is 840 × 2 + 3 = 1680 + 3 = 1683
  sum of digits = 1 + 6 + 8 + 3 = 18 divisible by 9
  so the required no. 1683
• 11 years agoHelpfull: Yes(0) No(0)