Find the smallest positive integer which ends in 17, is divisible by 17, and whose digits sum to 17.

Answer 1

clearly it can't be a 2 digit number, since it is to end in 17

make a list of multiple of 17 past 100
85, 102, 119, 136, 153, 170, 187, 204, 221, 238, 255, 272, 289, .... mhhh

ok, if it ends in 17 and the sum of the digits is also 17, the sum of the missing digits must be 9
leading 2 digits could be :
1817
2717
3617
4517
5417
6317
7217
8117
9017 none are divisible by 17

leading 3 digits add up to 9
10817
11717
12617
13517
14417
15317
16217
17117
18117
19017

20717
21617
...

Very tedious method, at this point it is all I can think of.
If I come up with an easier way, I will get back to you.
(Maybe Steve or Damon can see something that will make me say "duh, why didn't I think of that" )

Answer 2

Hey, I found one:

15317
is divisible by 17, ends in 17, and the sum of the digits is 17

I made up a very crude computer program using a computer language I taught in the 1980's , too complicated to explain here.

Answer 3

thanks