Question

How many integers are there such that $2\le n\le 100$ and the highest common factor of $n$ and $36$ is $1$?

• A. $166$
• B. $332$
• C. $331$
• D. $416$

Answer 1

The correct option is A $166$

${36.2}^2{.3}^2$

Since HCF (36,n) =1

Therefore, n sholud not be a multiple of 2 or 3.

$2\le n\le 100$

Total numbers = T= 1000.

Number of numbers divisible by 2= $N_2$

Number of numbers divisible by 3= $N_3$

Number of numbers divisible by 6= $N_6$

Therefore, there are $T-N_2-N_3+N_6$ total integers

a=2

l=1000

d=2

We know, $l=a+(N_2-1)d.$

$N_2=\frac{1000-2}{2}+1$

$N_2=\frac{998}{2}+1$

$N_2=499+1=500$

Similarly,

$N_3=\frac{999-3}{3}+1=333$

$N_6=\frac{999-6}{3}+1=166$

Answer

= 999-500-333+166

= 332