Question

Find the total number of integers ‘n’ such that $2\le n\le 2000$ and HCF of n and 36 is 1.

Answer 1

Hint: We will first find the numbers in the range $2\le n\le 2000$ which share a common divisor with 36, and then we will subtract it from total to get the answer.

Complete step-by-step answer:
So, to solve this question first we will find all the numbers divisible by 2 and divisible by 3 respectively. But as some integers are common to both 2 and 3, we will subtract them. The answer we get from this, is subtracted from the total to get the final answer.
In the first step, we will find the total number of integers divisible by 2. We know that 2,4,6, ….1998,200 are divisible by 2. Now we have to find the total number of these integers. As we know that the above series is A.P., we can find the total number of terms in the AP. Thus, $2000=2+\left( n-1 \right)2$ $\Rightarrow 1998=\left( n-1 \right)\left( 2 \right)$ $\Rightarrow n-1=1998\text{ }\Rightarrow n=1000$. Therefore, there are 1000 integers in $2\le n\le 2000$ which are divisible by 2.
In the second step, we will find the total number of integers divisible by 3. We know that 3,6,9, …..1995, 1998 are divisible by 3. Now we have to find the total number of these integers. As we know that the above series is AP, we can find the total number of terms in the AP.
Thus, $1998=3+\left( n-1 \right)3\Rightarrow 1995=\left( n-1 \right)3$
$\Rightarrow n-1=665$
$\Rightarrow n=666$
Therefore, there are 666 integers in $2\le n\le 2000$ which are divisible by 3.
Now we can see that there are integers which are divisible by both 2 and 3 such as 6,12,18,…1992,1998.
We have to subtract these numbers of terms from total. The total number of this terms is given by
$1998=6+\left( n-1 \right)6$
$1992=\left( n-1 \right)6$
$\Rightarrow n-1=332\text{ }\Rightarrow n=333$
Therefore, there are 333 integers in $2\le n\le 2000$ which are divisible by 6.
So, now total number of integers which are either divisible by 2 or 3 $=$ (number of integers divisible by 2) + (number of integers divisible by 3) $-$ (number of integer divisible by 6) $=\left( 666+1000 \right)-\left( 333 \right)=1333$
But we have to find the number of integers that are not divisible by 2 or 3. This can be obtained by subtracting the number of integers divisible by 2 or 3 from the total number of integers. Therefore,
Total integers $=1999-1333$
$=666$
So, the number of integers which have HCF $=$ 1 with 3 are 666.

Note: We can also obtain the number of integers divisible by 2 or 3 by finding the number of integers divisible by 6 and adding it to the number of integers which are odd multiples of 3. The final answer can be obtained by subtracting the sum from the total number of integers.