Subject: General Aptitude · Chapter: H.C.F. and L.C.M. of Numbers · Exam: · Marks: · Difficulty:
Find the least number which when divided by 16, 18, 20 and 25 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.
(a)17004
(b)18000
(c)18002
(d)18004
Answer
Answer (as printed): D
Explanation
L.C.M of 16, 18, 20, $25=3600$. Required number is of the form $3600 k+4$. Least value of $k$ for which $(3600 k+4)$ is divisible by 7 is $k=5$. $\therefore \quad$ Required number $=(3600 \times 5+4)=18004$.
Explanation as extracted from the printed page; notation may be imperfect.