Subject: General Aptitude · Chapter: H.C.F. and L.C.M. of Numbers · Exam: · Marks: · Difficulty:
A number $x$ is divided by 7 . When this number is divided by 8, 12 and 16. It leaves a remainder 3 in each case. The least value of $x$ is: [SSC-CHSL (10 + 2) Exam, 2015]
(a)148
(b)149
(c)150
(d)147
Answer
Answer (as printed): D
Explanation
LCM of 8, 12 and $16=48$ 2 | 8-12 - 16 ; 2 | 4-6 - 8 ; 2 | 2-3 - 4 ; 1-3-2 $$2 \times 2 \times 2 \times 2 \times 3=48$$ ∴ Required number $=48 a+3$ Which is divisible by 7. $$\begin{aligned} \therefore x=48 a+3 & =7 \times 6 a+6 a+3 \\ & =(7 \times 6 a)+(6 a+3) \text { which is divisible by } 7 . \end{aligned}$$ i.e. $6 a+3$ is divisible by 7 . When $a=3,6 a+3=18+3=21$ which is divisible by 7 . $$\therefore x=48 \times 3+3=144+3=147$$
Explanation as extracted from the printed page; notation may be imperfect.