The largest four digit number which when divided by 4,7 or 13 leaves a remainder of 3 in each case is?"

• To solve these kind of question always remember this formula,required number=
  n*(L.C.M. of a,b,c)+3 where n is a natural number,in this case n*(L.C.M. of 4,7, 13 )+3=n*364+3 now take n in such a way that this multiplication (n*364) should be closer to 9996,now you will wonder why 9996 let me explain 9996 because you have to add 3 more to 9996 so it will become 9999,but in our case that number is very less as compared to 9996 because if you will take n as 27 then (27*364) will become 9828,there is no point of getting 9996...anyway required number will be 9828+3=9831
• 11 years agoHelpfull: Yes(4) No(0)
• 4*7*13=364.
  multiply 364 with a num so that result is very near to 9999.
  hence 364*27=9828 and then add 3 to get remainder as 3.
  so, ans=9831.
• 11 years agoHelpfull: Yes(2) No(0)
• Answer is 9831
  EXPLANATION:
  L.C.M of 4,7 & 13=364
  Now if 9999/364 = 27.469 so it is clear that 364*27 is very near to 9999 which completely divide all the given three no(4,7 & 13)
  so finally our required Answer is 364*27 + 3 = 9831
• 11 years agoHelpfull: Yes(0) No(1)