How many two-digit numbers are divisible by 3?

Step 1: Identify the first two-digit number divisible by 3

The smallest two-digit number is 10. We check if 10 is divisible by 3. Since $10 = 3 \times 3 + 1$, it is not. The next multiple of 3 after 9 (which is a one-digit number) is $3 \times 4 = 12$. So, 12 is the first two-digit number divisible by 3.

Step 2: Identify the last two-digit number divisible by 3

The largest two-digit number is 99. We check if 99 is divisible by 3. Since $99 = 3 \times 33$, it is perfectly divisible by 3. Thus, 99 is the last two-digit number divisible by 3.

Step 3: Form an Arithmetic Progression (AP) and identify its components

The sequence of two-digit numbers divisible by 3 forms an Arithmetic Progression (AP) where the first term ($a$) is 12, the common difference ($d$) is 3, and the last term ($a_n$) is 99. We need to find the number of terms ($n$) in this AP.

Step 4: Apply the nth term formula for an AP

The formula for the $n^{th}$ term of an Arithmetic Progression is $a_n = a + (n-1)d$. We will substitute the values we found into this formula to solve for $n$.

Step 5: Substitute values and solve for n

Substituting $a=12$, $d=3$, and $a_n=99$ into the formula $a_n = a + (n-1)d$, we get $99 = 12 + (n-1)3$. Subtracting 12 from both sides gives $87 = (n-1)3$. Dividing by 3 yields $29 = n-1$. Finally, adding 1 to both sides gives $n=30$. Therefore, there are 30 such numbers.

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