Question

How many numbers of two digits are divisible by 5?

Answer 1

Step1: Calculation of $a,d$ and the last term.

Let $n$ numbers of two digits be divisible by 5.

All the 2 digit numbers divisible by 5 forms an A.P. series having a common difference $d=5$.

The smallest two-digit number divisible by 5 is 10.

Thus, the first term of this A.P. series is $a=10$.

The largest two-digit number divisible by 5 is 95.

Thus, the last or $n$th term of this A.P. series is $a_n=95$.

Step2: Calculation of the number of two-digit numbers divisible by 5.

The $n$th term of an A.P. is given by the formula $a_n=a+(n-1)d$, where $a$ is the first term and $d$ is the common difference.

Substitute the values $a_n=95$, $a=10$ and $d=5$ in equation $a_n=a+(n-1)d$ and solve for $n$.

$$\begin{gathered} 95=10+(n-1)5 \\ \Rightarrow 95-10=(n-1)5(Subtracting10frombothsides) \\ \Rightarrow 85=(n-1)5 \\ \Rightarrow \frac{85}{5}=n-1(Dividingbothsidesby5) \\ \Rightarrow 17=n-1 \\ \Rightarrow 17+1=n(Adding1tobothsides) \\ \therefore 18=n \end{gathered}$$

Final Answer: There exist 18 numbers of two digits that are divisible by 5.