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

This integer is x17, or xx17, or xxx17
all the digits make a sum of 17, ?+1+7=17, ?=9. the other digits need to make a sum of 9
of the numbers from 1-9, only 1 multiplying 7 will result in a 7 in the one's place, so my reasoning is that the ending 17 needs to remain independent, that is, we need to look for the first digits that will make a sum of 9 AND divisible by 17. Keep counting by 17, 
17, 34, 51, 68, 85, 102, 119, 136, 153
we can see that the first sum of 9 happens when 17*9=153, 1+5+3=9
so the smallest integer that satisfies all the conditions is 15317

Please let me know if you find another way to figure it out, or if there is a smaller interger

The right answer should be 6 but your not very specific