Subject: General Aptitude · Chapter: H.C.F. and L.C.M. of Numbers · Exam: 2007 · Marks: · Difficulty:
What is the greatest number of 3 digits which when divided by 6, 9 and 12 leaves a remainder of 3 in each case?
(a)903
(b)939
(c)975
(d)996
Answer
Answer (as printed): C
Explanation
Greatest number of 3 digits is 999. L.C.M. of 6, 9 and 12 is 36. On dividing 999 by 36, the remainder obtained is 27 . So, required number $=(999-27)+3=975$.
Explanation as extracted from the printed page; notation may be imperfect.