Question 1108009
<br>
My first inclination, whenever I see a problem involving divisibility by 11, is to use the rule for divisibility by 11.  That rule, applied to this problem, would say that either
(1) (9+b+2)-(a+c) = 0, or
(2) (9+b+2)-(a+c) = 11, or
(3) (9+b+2)-(a+c) = -11<br>
Logical analysis shows there are 55 solutions of type (1) and 36 of type (2) for a total of 91; there are no solutions of type (3).<br>
Completing the logical analysis to reach the answer by that method is good mental exercise.<br>
However, for this particular problem, there is a much faster path to the answer.<br>
The first and last digits are fixed; and the number 90002 is divisible by 11.<br>
Since the last digit is fixed, to get from one number divisible by 11 to the next, you have to add 110.<br>
So the answer to the problem is the number of times you can add 110 to 90002 and still have 9 as the leading digit.<br>
{{{99999-90002 = 9997}}}
{{{9997/110 = 90.88}}} (approximately)<br>
That calculation shows there are 90 more numbers after 90002 with leading digit 9 and last digit 2 that are divisible by 11; so the total number of numbers with those characteristics is 91.