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

How many two-digits numbers are divisible by 3

Solution;

The two-digit numbers divisible by 3 are 12, 15, 18, …, 99. Clearly, these number are in AP.

Here, a = 12 and d =15 —12 = 3

Let this AP contains n terms. Then,

an = 99

12 + (n – 1) x 3=99         [an=a+(n —1)cl]

3n-F9=99

3n = 99-9 = 90

n = 30

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

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

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Answer

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