Subject: General Aptitude · Chapter: Number System · Exam: 2016 · Marks: · Difficulty:
The number of three digit numbers which are multiples of 9 are
(a)100
(b)99
(c)98
(d)101
Answer
Answer (as printed): A
Explanation
The first 3-digit number which is divisible by 9 is 108 and last three digit number which is divisible by 9 is 999 . So, we have an AP with $a=108, d=9$ and $a_{n}=999$ $$\begin{aligned} & \therefore a_{n}=a+(n-1) d \\ & \Rightarrow 999=108+(n-1) 9 \\ & \Rightarrow 999-108=(n-1) 9 \\ & \Rightarrow 891=(n-1) 9 \\ & \Rightarrow(n-1)=\frac{891}{9}=99 \\ & \Rightarrow n+99+1=100 \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.