>>11593799>>11593987the picture asks you to compute 9n+9 = 9(n+1) and sum the digits. no matter what number you choose, you will always end up with a multiple of 9.
so let's see what needs to happen to end with 18. we need a number whose sum of digits is 18 and at the same time is a multiple of 9 (because the algorithm produces only multiples of 9). smallest such number is 99. now you're looking for a solution to
99 = 9n+9
which gives n = 10.
why did I guess that 99 satisfies the criterion? because 9+9 = 18 and this is the only possibility with just 2 summands. you can write for example 18 = 9 + 8 + 1. now, because of the criterion for divisibility by 9, we know that all permutations of 9.8,1 give a three digit number which is divisible by 9. we obtain the equations
189 = 9n + 9
198 = 9n + 9
819 = 9n + 9
891 = 9n + 9
918 = 9n + 9
981 = 9n + 9
where the first two equations give n = 20, 21.
and so on.
