Question 5

We can solve this problem using the properties of an arithmetic progression.

Step 1 — Count the numbers

Let's list the 2-digit numbers divisible by 3. The first 2-digit number divisible by 3 is 12. The last 2-digit number divisible by 3 is 99. These numbers form an arithmetic progression. The first term, $a$, is 12. The common difference, $d$, is 3. The last term, $t_n$, is 99. We use the formula $t_n = a + (n-1)d$.

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

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

$87 = 3(n - 1)$

$\frac{87}{3} = n - 1$

$29 = n - 1$

$n = 29 + 1$

$\boxed{n = 30}$

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

Step 2 — Calculate the sum

Now, let's find the sum of these 30 numbers. We use the sum formula $S_n = \frac{n}{2} \times (a + l)$. Here, $n$ is 30, $a$ is 12, and $l$ is 99.

$S_{30} = \frac{30}{2} \times (12 + 99)$

$S_{30} = 15 \times 111$

$\boxed{S_{30} = 1665}$

The sum of all these numbers is 1665.

Answer

(i) Number of 2-digit numbers divisible by 3 = 30 (ii) Sum of all these numbers = 1665