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What's the best way to find if a large number, with all non-zero digits, in the real numbers is divisible by 3 or 6?

Interested in this because of how hard it seems to find if a large number with all non-zero digits is divisible by 3 or not. I mean I could be wrong but from my first examination, it seems like for very large numbers (I'm talking more than 9 digits certainly), when trying to use the method of summing the digits and testing their divisibility by 3, you could easily wind up with a chain of sums that has to be resolved continually into smaller and smaller numbers until you find one which you can see has a sum divisible by 3.

Any easier test or formula we could use for very large numbers?

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    $\begingroup$ Easier than sum of digits? This doesn't make sense. $\endgroup$ Commented Jun 25, 2021 at 22:13
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    $\begingroup$ What exactly do you mean by "large"? Would it take less than a minute to write down the digits? That's not large in this context. "It barely fits in the observable universe" will begin to scratch the surface of "large", but even then, three repetitions of digit sums will take it down to manageable levels. $\endgroup$ Commented Jun 25, 2021 at 22:24
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    $\begingroup$ And how are these numbers represented when they are given to you? Are you just given the base-ten digit representation, or are they presented in some other manner? $\endgroup$ Commented Jun 25, 2021 at 22:32
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    $\begingroup$ When summing digits, you don't have to sum them as integers, instead sum them $\bmod 3$. $\endgroup$ Commented Jun 25, 2021 at 22:45
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    $\begingroup$ @Alan According to this link, that number is equal to $$2\uparrow^{g_{64}-2}(g_{64}+3)-3=2^{2^{\cdot^{\cdot^{\cdot^2}}}}-3$$We see that this is clearly not divisible by $3$, as any even power of $2$ is a power of $4$, and therefore of the form $3k+1$. (That number is also not really that large. Most natural numbers are, in fact, significantly larger.) $\endgroup$ Commented Jun 25, 2021 at 23:34

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Divisibility by 3 rule: Add up all of the digits of the number, and if that number is, then that original number is divisible by 3. If you still get a very large number, repeat the rule for that number. Do this until you get a "nice" number, that can be easily seen if divisible by 3. This is the easiest method.

Divisibility by 6: First thing is, that number must be even, or else it is not divisible by 6. If the number is even, then apply the divisibility by 3 rule (see above).

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  • $\begingroup$ This should be what you are looking for. $\endgroup$ Commented Jun 26, 2021 at 14:58

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