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Little math problem(s) December 2018 - pier4r - 12-19-2018 08:54 PM
Given a random positive integer N, compute the sum S of its digits (for example 50 -> S = 5 + 0 = 5). Subtract this sum from N, thus getting the result "R=N-S".
What is the smallest divisor greater than 1 that can divide evenly every R so computed? Why is it so?
RE: Little math problem December 2018 - Thomas Klemm - 12-19-2018 09:20 PM
Code:
Spoiler Alert!
Since 10^k - 1 ≡ 0 (mod 9) the result R is divisible by 9.
The smallest divisor of 9 greater than 1 is 3.Cheers
Thomas
RE: Little math problem December 2018 - pier4r - 12-19-2018 10:39 PM
Thanks for the spoiler. I completely forgot to give a template.