A number theory problem by Abhinav SSingh

We check for multiples of 7,

$105,112, \cdots , 490,497$

Using formula for $n^{\text{th}}$ term of an AP, the number of these multiples are

$n_1 = 1 + \dfrac{497 - 105}{7} = 57$

Checking for multiples of 21,

$105,126,\cdots,462,483$

The number of these multiples are

$n_2 = 1 + \dfrac{483 - 105}{21} = 19$

Since all the multiples of 21 are also included in the set of multiples of and we only want those multiples of 7 which are not multiples of 21 so our required answer is $n_1 - n_2 = \boxed{38}$ .

500÷7=71.428 (means 71 whole multiple of 7 under 500) 100÷7=14.28 (means 14 whole under 100) Difference =57 i.e. 57 multiples of 7 are between 100 and 500. Similarly for 21 we find out that ther are 19 multiples between 100 & 500. Therefore numbers divisible by 7 but not by 21 =57-19=38