Question

Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.

Answer 1

Total three digit numbers $=900$ [ from $100$ to $999$]

Number of three digit numbers divisible by $3=300$

Probability that the number written by the student is divisible by $3$

$=\frac{\text{Number of three-digit numbers divisible by 3}}{\text{Total three-digit numbers}}$

$=\frac{300}{900}=\frac{1}{3}$

Way to find total 3 digit numbers divisible by 3:

First 3 digits number divisible by $3=102$

so the series will be $102,105,108,.........999$

Last term which is divisible by $3=999$

The above series is in AP.

We know the $n^{th}$ of an AP is given by

$a_n=a+(n-1)d$ [Here, $a=$ first term $=102,$ $d=$common difference $=105-102=3$]

$999=102+(n-1)3$ [Let $a_n=999$]

$\Rightarrow 299=n-1$

$\therefore n=300$

there are $300$ three digit numbers which are divisible by $3$.