Subject: General Aptitude · Chapter: H.C.F. and L.C.M. of Numbers · Exam: 2004 · Marks: · Difficulty:
What is the third term in a sequence of numbers that leave remainders of 1, 2 and 3 when divided by 2, 3 and 4 respectively?
(a)11
(b)17
(c)19
(d)35
Answer
Answer (as printed): D
Explanation
Clearly, $(2-1)=1,(3-2)=1$ and $(4-3)=1$. L.C.M. of 2, 3, 4 = 12. So, the sequence shall have numbers of the form $12 k-1$, where $k=1,2,3, \ldots \ldots$. ∴ Third term of the sequence $=12 \times 3-1$ $$=36-1=35 .$$
Explanation as extracted from the printed page; notation may be imperfect.