What is the smallest two digit number divisible by 3 and 9?

term of an arithmetic progression. Then, we will take the required smallest 3-digit number as the first term, the required largest 3-digit number as the last term and 9 as the common difference of the A.P. We will find the number of terms of the A.P. and the number of terms will be the number of 3-digit numbers divisible by 9.

Formulas used: We will use the formula, l=a+(n−1)d

where l

is the last term, d

is the common difference and n

is the total number of terms in an A.P.

Complete step-by-step answer:

We know that the divisibility rule of 9 is given as: if the sum of digits of a number is divisible by 9, the number is also divisible by 9.

Let’s go through the multiples of 9 and find the smallest 3-digit number that is a multiple of 9.

We know that

9×11=999×12=108

We have found that 108 is the smallest 3-digit number that is divisible by 9.