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Hint- Here, we will be using the formula for ${n^{{\text{th}}}}$ term or last term of an arithmetic progression.

Since, we know that the numbers which will be divisible by both 2 and 5 will be the numbers which will be divisible by 10.

Now we have to find the natural numbers between 101 and 999 which are divisible by 10

So, the first natural number between 101 and 999 which is divisible by 10 is 110 and the last natural number between 101 and 999 which is divisible by 10 is 990.

So, the natural numbers between 101 and 999 which are divisible by 10 are 110,120,130, …. ,980, 990.

Clearly, we can see that the above series is an arithmetic progression with the common difference of 10.

As we know that for any arithmetic progression with first term as ${a_1}$, common difference as $d$, number of terms in the series as $n$ then the last term ${a_n}$ is given by ${a_n} = {a_1} + \left( {n - 1} \right)d{\text{ }} \to {\text{(1)}}$

For the given AP, ${a_1} = 110$, $d = 10$ and ${a_n} = 990$

Using formula given by equation (1), we get

${a_n} = {a_1} + \left( {n - 1} \right)d \Rightarrow 990 = 110 + 10\left( {n - 1} \right) \Rightarrow 880 = 10\left( {n - 1} \right) \Rightarrow \left( {n - 1} \right) = 88 \Rightarrow n = 89$

Hence, the total number of natural numbers which are divisible by both 2 and 5 (or divisible by 10) between 101 and 999 are 89.

Note- In this particular problem, the numbers which will be divisible by both 2 and 5 will be the numbers which will have both 2 and 5 as their prime factors i.e., the number should be a multiple of 10. Here, the first number and the last number which is divisible by both 2 and 5 can be easily predicted.

Since, we know that the numbers which will be divisible by both 2 and 5 will be the numbers which will be divisible by 10.

Now we have to find the natural numbers between 101 and 999 which are divisible by 10

So, the first natural number between 101 and 999 which is divisible by 10 is 110 and the last natural number between 101 and 999 which is divisible by 10 is 990.

So, the natural numbers between 101 and 999 which are divisible by 10 are 110,120,130, …. ,980, 990.

Clearly, we can see that the above series is an arithmetic progression with the common difference of 10.

As we know that for any arithmetic progression with first term as ${a_1}$, common difference as $d$, number of terms in the series as $n$ then the last term ${a_n}$ is given by ${a_n} = {a_1} + \left( {n - 1} \right)d{\text{ }} \to {\text{(1)}}$

For the given AP, ${a_1} = 110$, $d = 10$ and ${a_n} = 990$

Using formula given by equation (1), we get

${a_n} = {a_1} + \left( {n - 1} \right)d \Rightarrow 990 = 110 + 10\left( {n - 1} \right) \Rightarrow 880 = 10\left( {n - 1} \right) \Rightarrow \left( {n - 1} \right) = 88 \Rightarrow n = 89$

Hence, the total number of natural numbers which are divisible by both 2 and 5 (or divisible by 10) between 101 and 999 are 89.

Note- In this particular problem, the numbers which will be divisible by both 2 and 5 will be the numbers which will have both 2 and 5 as their prime factors i.e., the number should be a multiple of 10. Here, the first number and the last number which is divisible by both 2 and 5 can be easily predicted.

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