LogFAQs > #885653319

Eevee-Trainer posted...

A digital sum of 3, 6, or 9 by the end one could argue only make up 30% of the outcomes because the end result would be in the range of integers 0-9 inclusive. Though only 0 has a digital sum of zero so in the infinite set of integers ignoring zero - to give nine outcomes on 1-9 - would be more reasonable, yielding a proportion of 1/3. I could this out practically in an Excel sheet perhaps but that'd have to wait until morning. I don't see why what you're suggesting be a thing though beyond confirmation bias.

Let's assume that the end result is 3/6/9. Consequentially the number they were summed from is a multiple of 3 (since multiples of 3 have digital sums that are a multiple of 3). And likewise this continues all of the way back to the original product, implying the original sum is a multiple of 3.

Thus this occurs 33.33...% of the time since I believe a good measure for the numbers that are evenly divisible by a number n is 1/n (here n = 3). 1/3 of numbers are multiples of 3 and therefore 1/3 of numbers have a digital sum of 0.

That sounded like the implication of what you originally said but I get that result more than1/3 of the time. I'd say slightly more than half. Maybe even 3/4. Best way to test would be to do it with a bunch of random numbers and see how often the digital sum is 3,6, or 9.
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