Subject: General Aptitude · Chapter: H.C.F. and L.C.M. of Numbers · Exam: 2005 · Marks: · Difficulty:
What is the least number which when divided by the numbers 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but when divided by 13 leaves no remainder?
Answer
Answer (as printed):
Explanation
L.C.M. of 3, 5, 6, 8, 10 and 12 = 120. So, the required number is of the form 120 k + 2. Least value of k for which (120k + 2) is divisible by 13 is k = 8. \ Required number = (120 × 8 + 2) = 962.
Explanation as extracted from the printed page; notation may be imperfect.