Question

Find the number of integers between 100 and 1000 that are
(a). divisible by 7
(b). not divisible by 7

Answer 1

Hint: Here, in this given question, we can see that every consecutive number that is divisible by 7 has a common difference of 7. So, all the numbers that we will get between 100 and 1000 that are divisible by 7 will form an Arithmetic Progression (AP). We just need to find out the first term and the last term and then the number of terms in the AP to get our first answer. Then we can subtract it from 899 to get our second answer as there are 899 numbers between 100 and 1000.

Complete step-by-step answer:

In this given question, we are asked to find out the number of integers between 100 and 1000 that are divisible by 7 the number of terms which are not divisible by 7.
We can see that every consecutive number that is divisible by 7 has a common difference of 7. So, all the numbers that we will get between 100 and 1000 that are divisible by 7 will form an Arithmetic Progression (AP).
Now, the first term of the AP can be found out as following:
$\begin{aligned} & \dfrac{100}{7}=14\dfrac{2}{7} \\\ & \Rightarrow 100=14\times 7+2 \\\ \end{aligned}$
So, $14\times 7=98$ is the last number that is divisible by hundred and less than 100.
So, $98+7=105$ is the first multiple of 7 between 100 and 1000. Therefore, 105 is the first term of the AP…………………(1.1)
Now, for the last term, we do the following:
$\begin{aligned} & \dfrac{1000}{7}=142\dfrac{6}{7} \\\ & \Rightarrow 1000=142\times 7+6 \\\ \end{aligned}$
So, $142\times 7=994$ is the last number with a multiple of seven and is less than 1000.
Therefore, the last term of our AP is 994.
Now, for the nth term of an AP $\left( {{a}_{n}} \right)$ the formula is $\left( {{a}_{n}} \right)=a+\left( n-1 \right)d...........(1.1)$, where a is the first term and d is the common difference.
Here, $\left( {{a}_{n}} \right)$, a and d are 994, 105 and 7 respectively. So, putting these values in 1.1, we get,
$\begin{aligned} & 994=105+\left( n-1 \right)\times 7 \\\ & \Rightarrow \dfrac{994-105}{7}=n-1 \\\ & \Rightarrow 127+1=n \\\ & \Rightarrow n=128..............(1.2) \\\ \end{aligned}$
Hence, we get the no. of terms of the AP as 128.
So, 128 numbers between 100 and 1000 are divisible by 7.

(b). Now, there are 899 numbers between 100 and 1000.
So, $899-128=771$ numbers are not divisible by 7 between 100 and 1000.
Therefore, we have got the first answer as 128 and the second one as 771.

Note: We should be careful to add 7 to get the first term divisible by 7 between 100 and 1000, it is because upon dividing 100 by 7, the remainder was 2 and was subtracted from 100, thus the obtained number, 98 was less than 100 and hence did not lie between 100 to 1000.