Find a positive integer that ends in 17, is divisible by 17 and the sum of its digits is equal to 17.
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Find a positive integer that ends in 17, is divisible by 17 and the sum of its digits is equal to 17.
15,317...
- ends in 17
- divisible by 17 (15317/17 = 901)
- sum of digits is 17 (1 + 5 + 3 + 1 + 7 = 17)
However, I found this by guess-and-check. I also don't know if this is the smallest integer possible.
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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.Quote:
Originally Posted by souksfxc
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.
Hi HallsofIvy,
Thank you for your comprehensive solutions.
SoUkSFXC
Hi HallsofIvy,
The sum of digiits are not equal to 17 when m=27, 45 and 54.
Am I doing the calculation wrong here?
souksfxc
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.
01