The number of numbers from 1 to 200 which are divisible by neither 3 nor 7 is?

Answer: You can do this by A.P.

First, let’s see how many numbers are divisible by 3.

It goes like this: 3,6,9,12.......198

By using the formula, we get, a+(n-1) d = an; 3+(n-1)3 = 198

n=66

So the total no. of numbers that are divisible by 3 is 66.

Now the numbers that are divisible by 7.

It goes like, 7,14,21....196

Use the formula again a+(n-1) d = an; 7+(n-1)7 = 196

n=28

So the total no. of numbers that are divisible by 7 is 28.

Now the numbers that are divisible with both 3 and 7.

It goes like: 21, 42, 63....189

Use the formula a+(n-1) d = an; 21+(n-1)21 = 189

n=9

So the total no. of numbers that are divisible by both 3 and 7 are 9.

Finally, from 1 to 200, there are 200 numbers. We have to remove those numbers that are either divisible by 3 or 7. Here it goes, 200 - (66 + 28) = 106.

Now certain numbers are repeated in both 3 and 7, like 21,42,63 etc. These numbers have been removed twice. So we can add these numbers once. So the final answer is 106 + 9 = 115. Hope it helps!