Subject: General Aptitude · Chapter: H.C.F. and L.C.M. of Numbers · Exam: · Marks: · Difficulty:
Find the least multiple of 23 , which when divided by 18, 21 and 24 leaves remainders 7, 10 and 13 respectively.
(a)3002
(b)3013
(c)3024
(d)3036
Answer
Answer (as printed): B
Explanation
Here $(18-7)=11,(21-10)=11$ and $(24-13)=11$. L.C.M. of 18, 21 and 24 is 504. Let the required number be $504 k-11$. Least value of $k$ for which $(504 k-11)$ is divisible by 23 is $k=6$. $\therefore \quad$ Required number $=504 \times 6-11=3024-11=3013$.
Explanation as extracted from the printed page; notation may be imperfect.