# A question in division

• Jun 13th 2009, 08:58 PM
souksfxc
A question in division
Find a positive integer that ends in 17, is divisible by 17 and the sum of its digits is equal to 17.
• Jun 13th 2009, 10:19 PM
yeongil
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.

01
• Jun 14th 2009, 04:26 AM
HallsofIvy
Quote:

Originally Posted by souksfxc
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.
• Jun 15th 2009, 01:06 AM
souksfxc
division
Hi HallsofIvy,

Thank you for your comprehensive solutions.

SoUkSFXC
• Jun 15th 2009, 01:20 AM
souksfxc
Division by 17
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
• Jun 15th 2009, 01:33 AM
yeongil
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