Let the number be 17n=100a+b because there is no tens digit and a+b=9, so b=9-a.
17n=100a+9-a=99a+9=9(11a+1)≤999. 11a+1≤111, so 11a≤110, a<10.
If a=3 11a+1=34 which is divisible by 17. So 17n=9×34=306. This is the only solution because a cannot exceed 9.