Question 1029343
The highest three-digit number divisible by 21 is 987, while the smallest is 105.
Use the formula for the nth term of an AP 

{{{a[n] = a[1] + (n-1)d}}}, where {{{a[n] = 987}}} and {{{a[1] = 105}}}, and d = 21.

==> 987 = 105 + (n-1)21 ==> 882 = 21(n-1)  ==> 42 = n-1 ==> n = 43.

Thus there are 43 such numbers.


Note: I know there is another, much shorter (or more elegant) method in doing this, and expect another tutor to post it.