# proof

• May 30th 2005, 05:18 AM
Alex F.
proof
How can I prove that 153 is the only number for which "number = 17 * sum of the number's digits" (153 = 17*(1+5+3)) is true?

Thanks :)
• Jun 5th 2005, 11:33 PM
beepnoodle
Suppose that there is another solution, call it x. Show that x must be equal to 153. This shows that your solution is unique.
• Jun 6th 2005, 05:32 AM
hpe
Quote:

Originally Posted by Alex F.
How can I prove that 153 is the only number for which "number = 17 * sum of the number's digits" (153 = 17*(1+5+3)) is true?

Thanks :)

First off, x must evidently be divisible by 17, x = 17n.
Also, x cannot have more than three digits, x = abc, because otherwise n = x/17 would be larger than 1000/17 > 58. Then there would have to be at least seven digits (because each digit contributes at most 9 to this sum), so x/17 would have to be larger than 10,000,000/17 > 580000. Therefore there would have to be at least 70,000 digits, and so on, a contradiction.

Suppose now x has the digits abc, that is x = c+10b+100a.
Then you want also x = 17*(a+b+c), and a, ..., g must be integers from the list {0,1,2,...,9}. Work with this to prove that a=1, b=5, c=3.