Question

How many two-digit numbers are divisible by both 2 and 3?

Answer 1

Hint: Here, we will first find the total number of two-digit numbers. Then we will write down a few numbers that are divisible by both 2 and 3 and try to find the pattern or relation between each number. Then observing the pattern, we will apply the formula of $$n^{th}$$ term of an AP and simplify further to get the required answer.

Complete step by step solution:
We know that two-digit numbers lie between 10 and 99.
First-term that is divisible by 3 is 12 which is also divided by 2 then we have 18 which is divisible by 2 and 3 both then we have 24 which is again divisible by 2 and 3.
So, the numbers divisible by both 2 and 3 are $$12,18,24,......96$$.
As we can see that the above relation is an A.P with first term as 12, common difference as 3 and $$n^{th}$$ term as 96.
Substituting $$a=12,d=6$$ and $$T_n=96$$in the formula $$n^{th}$$ term of AP $$T_n=a+d\left(n-1\right)$$, we get
$$96=12+6\left(n-1\right)$$
Multiplying the terms, we get
$$\Rightarrow 96=12+6n-6$$
Rewriting the equation, we get
$$\Rightarrow 6n=96-12+6$$
Adding and subtracting the like terms, we get
$$\Rightarrow 6n=90$$
Dividing both side by 6, we get
$$\Rightarrow n=15$$

Therefore, we can conclude that there are 15 two-digits numbers that are divisible by both 2 and 3.

Note:
The divisibility rule is a method by which we can easily find whether a number is divisible by a particular number or not. We can only say that to find numbers divisible by 2 and 3 both we follow divisibility of 6 or we can say if a number is divisible by 2 and 3 it also satisfies the divisibility of digit 6. Here, the numbers that are divisible by both 2 and 3 form an AP series. AP or an arithmetic progression is a series or sequence in which the consecutive numbers differ by a common difference.