Subject: General Aptitude · Chapter: H.C.F. and L.C.M. of Numbers · Exam: 2005 · Marks: · Difficulty:
Find the greatest number of 4 digits which when divided by 4, 5, 6, 7 and 8 leaves 1, 2, 3, 4 and 5 as remainders.
(a)9237
(b)9240
(c)9840
(d)9999
Answer
Answer (as printed): A
Explanation
Clearly, $(4-1)=3,(5-2)=3,(6-3)=3,(7-4)=3$ and $(8-5)=3$. L.C.M. of 4, 5, 6, 7, $8=840$. Greatest number of 4 digits $=9999$. On dividing 9999 by 840, the remainder is 759 . So, required number $=(9999-759)-3=9237$.
Explanation as extracted from the printed page; notation may be imperfect.