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

Last updated date: 11th Aug 2024
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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.