How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?           [Page No. 186]

Two-digit multiples of 3 form the A.P. 12, 15, 18, ..., 99.

$$a = 12, d = 3, \text{ last term } l = 99$$

Number of terms:

$$99 = 12 + (n - 1)3$$

$$n = \frac{(99 - 12)}{3} + 1$$

$$n = \frac{87}{3} + 1 = 29 + 1 = 30$$

So, there are total 30 two-digit numbers that are divisible by 3.

Now, the required sum is given by:

$$S = 12 + 15 + 18 + \dots + 99$$

Taking 3 common from each term, we get:

$$S = 3(4 + 5 + 6 + \dots + 33)$$

$S = 3 \left[ (1 + 2 + 3 + \dots + 33) - (1 + 2 + 3) \right]$

Using the formula for the sum of the first $n$ natural numbers, $1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$,

we have:

$$S = 3\left[\frac{(33 \times 34)}{2} - \frac{(3 \times 4)}{2}\right]$$

$$S = 3[561 - 6]$$

$$S = 3 \times 555$$

$$S = 1665$$

Hence, there are 30 two-digit numbers divisible by 3, and the sum of all these numbers is 1665.