Find a positive integer that ends in 17, is divisible by 17 and the sum of its digits is equal to 17.
If it is divisible by 17 it must be of the form 17n for some integer n. If it ends in 17, then n must be of the form 100m+ 1 for some integer m. 17 itself has digit sum 8 so m must have digit sum 17- 8= 9. The smallest such integer is, of course, 9. m= 9, n= 901 and 17n= 15317. yeongil's answer is, in fact, the smallest possible answer.
18 also has digit sum 9 so m= 18, n= 1801 and17n= 30617 is the next larger answer.
You can keep generating answers by taking m= 27, 36, 45, 54, 63, and 72, the multiples of 9 up to 8*9= 72. If you take m= 9*9= 81, n= 100m+1= 8101 and 17n= 137717 which has digit sum 17+ 9= 26.
Those are the only solutions.
No, you're not wrong. The digits add up to 26 when m = 27, 45, or 54. I've also found more solutions that work.
m = 9 -> 15,317
m = 18 -> 30,617
m = 36 -> 61,217
m = 63 -> 107,117
m = 72 -> 122,417
m = 90 -> 153,017
m = 126 -> 214,217
m = 153 -> 260,117
m = 180 -> 306,017
m = 189 -> 321,317
m = 243 -> 413,117
There are more, I'm sure.
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