What is the greatest number of 4 digits that when divided by any of the numbers 6, 9, 12, 17 leaves a remainder of 2?
1) 9792
2) 9982
3) 9794
4) 9906
5) None of these

• take L.C.M of all numbers i.e 6,9,12,17
  the L.C.M is 612 therefore 612 is divisible by all numbers
  so multiple of 612 will also be divisible by all the above numbers
  so 612*17=9792 which is divisible by all
  add 2 to 9792 to get a remainder of 2 as required in the question..
  so ans is 9792+2=9794
• 7 years agoHelpfull: Yes(6) No(2)
• Soln to find n-digit greatest number divided by a, b, c leaving same remainder K
  
  https://books.google.co.in/books?id=CIzCCwAAQBAJ&lpg=SA2-PA4&ots=2CwuyFvMc1&dq=n-digit%20greatest%20number%20divided%20by%20a%2C%20b%2C%20c&pg=SA2-PA4#v=onepage&q&f=false
  
  9999/612
  Q=16 R=207
  ans = 9999-207+2
  =9794
  
• 7 years agoHelpfull: Yes(4) No(3)
• Apply hit and trial method at each option
  9792/9= 0 remainder (failed)
  9982/9= 1 remainder (failed)
  9794/6= 2
  9794/9= 2
  9794/12=2
  9793/17= 2
  condition is satisfy so answer is option c
• 7 years agoHelpfull: Yes(3) No(0)
• Take LCM of all the numbers 6,9,12,17 =612
  and devide each option
  (3) 9794
• 7 years agoHelpfull: Yes(1) No(0)