Question

How many numbers are lying between 100 and 500 that are divisible by 7 and not by 21?
(a) 57
(b) 19
(c) 38
(d) None of these

Answer 1

Hint: We start solving the problem by dividing 100 with 7 and finding the quotient and remainder. We can see that it is not divisible by 7 and gives a remainder between 1 and 6. We then multiply the succeeding number of quotient with 7 to get the first term required. We then divide 500 and subtract it with the remainder to the final term of the numbers. We then use $n^{th}$ term of arithmetic progression to get the total numbers of terms that were divisible by 7. Similarly, we do it for terms that were divisible by 21. We then subtract the total number of terms that were divisible by 7 with the total number of terms that were divisible by 21 to get the desired result.

Complete step-by-step solution:
According to the problems, we need to find how many numbers are present between 100 and 500 which were divisible by 7 but not by 21.
We know that 21 is divisible by 21. So, all the terms that were multiples of 21 are divisible by 7. In order to get the terms that were only divisible by 7, we subtract the total terms divisible by 21 from total terms divisible by 7.
Let us first divide 100 with 7.
\[\begin{align}
  & \left. 7 \right)100\left( 14 \right. \\
 & \underline{\text{ 7 }} \\
 & \text{ 30} \\
 & \underline{\text{ 28 }} \\
 & \text{ 2} \\
\end{align}\].
So, we can see that the remainder is 2 and 100 is 2 greater than the number that is exactly divisible by 7. But It will be less than 100, we also got a quotient as 14. So, our first number which is divisible by 7 and greater than 100 will be 15 times 7.
The first number that is greater than 100 and divisible by 7 is $\left( 7\times 15 \right)=105$. We can get the next numbers by adding 7 to the preceding terms.
So, we get the terms like 105, 112, 119,………. . Let us find the last term of the series by dividing 500 with 7.
So, let us divide 500 with 7.
\[\begin{align}
  & \left. 7 \right)500\left( 71 \right. \\
 & \underline{\text{ 49 }} \\
 & \text{ 10} \\
 & \underline{\text{ 7 }} \\
 & \text{ 3} \\
\end{align}\].
So, we can see that the remainder is 3 and 500 is 3 greater than the number that is exactly divisible by 7. So, to get the number which is less than 500 and divisible by 7 we subtract 500 with 3.
The last number that is present between 100 and 500 which is divisible by 7 is $\left( 500-3 \right)=497$.
So, we get the numbers that are lying between 100 and 500 and that is divisible by 7 as
105, 112, 119, ……, 490, 497 ---(1).
We can see that the numbers in (1) follow the arithmetic progression (A.P) with first term 105 and common difference 7. We need to find the total no. of terms in (1). So, we use the nth term formula of A.P (arithmetic progression).
We know that the nth term of an arithmetic progression is defined as ${{T}_{n}}=a+\left( n-1 \right)d$, where ‘a’ is the first term and ‘d’ is a common difference.
So, we have $497=105+\left( n-1 \right)7$.
$\Rightarrow 497=105+7n-7$.
$\Rightarrow 497=98+7n$.
$\Rightarrow 399=7n$.
$\Rightarrow n=\dfrac{399}{7}$.
$\Rightarrow n=57$.
So, we have a total of 57 terms lying between 100 and 500 that were divisible by 7.
Let us find the total no. of terms lying between 100 and 500 that were divisible by 21.
Let us first divide 100 with 21.
\[\begin{align}
  & \left. 21 \right)100\left( 4 \right. \\
 & \underline{\text{ 84 }} \\
 & \text{ 16} \\
\end{align}\].
So, we can see that the remainder is 16 and 100 is 16 greater than the number that is exactly divisible by 21. But It will be less than 100, we also got a quotient as 4. So, our first number which is divisible by 21 and greater than 100 will be 5 times of 21.
The first number that is greater than 100 and divisible by 21 is $\left( 21\times 5 \right)=105$. We can get the next numbers by adding 21 to the preceding terms.
So, we get the terms like 105, 126, 147,………. . Let us find the last term of the series by dividing 500 with 21.
So, let us divide 500 with 21.
\[\begin{align}
  & \left. 21 \right)500\left( 2 \right.3 \\
 & \underline{\text{ 42 }} \\
 & \text{ 80} \\
 & \underline{\text{ 63 }} \\
 & \text{ 17} \\
\end{align}\].
So, we can see that the remainder is 17, and 500 is 17 greater than the number that is exactly divisible by 21. So, to get the number which is less than 500 and divisible by 21 we subtract 500 with 17.
The last number that is present between 100 and 500 which is divisible by 7 is $\left( 500-17 \right)=483$.
So, we get the numbers that are lying between 100 and 500 and that is divisible by 21 as
105, 126, 147, ……, 462, 483 ---(2).
We can see that the numbers in (2) follow the arithmetic progression (A.P) with first term 105 and common difference 21. We need to find the total no. of terms in (2). So, we use the nth term formula of A.P (arithmetic progression).
We know that the nth term of an arithmetic progression is defined as ${{T}_{n}}=a+\left( n-1 \right)d$, where ‘a’ is the first term and ‘d’ is a common difference.
So, we have $483=105+\left( n-1 \right)21$.
$\Rightarrow 483=105+21n-21$.
$\Rightarrow 483=84+21n$.
$\Rightarrow 399=21n$.
$\Rightarrow n=\dfrac{399}{21}$.
$\Rightarrow n=19$.
So, we have a total of 19 terms lying between 100 and 500 that were divisible by 21.
Since we need the total no. of terms lying between 100 and 500 that were divisible by 7 but not by 21.
So, we subtract 57 terms of numbers that were divisible by 7 with 19 terms of numbers that were divisible by 21.
Total no. of terms = $\left( 57-19 \right)=38$.
We have found a total of 38 numbers lying between 100 and 500 that were divisible by 7 but not 21.
The correct option for the given problem is (c).

Note: We can also solve the problem by writing the terms in order and manually counting the terms but it will take a lot of time to do it and requires high attention. We should not make mistakes while taking the nth term of an arithmetic progression. We need to make sure that all multiples of 21 are divisible by 7 but the converse is not true. Similarly, we can expect problems to find the total number of terms that were divisible by 3 but not by 7.