-1

We know that if we add up all the digits and the sum is divisible by $3$, then the whole number is divisible by $3$. My question is:

Are there any other methods that a number is divisible by 3, or is the sum of digits the only one known?

Is there a possibility that another method exists?

Thanks

Greg Martin
  • 78,820
user356448
  • 47
  • The same idea works for 9. – herb steinberg Aug 08 '21 at 04:06
  • Well, you can write the number in base 3 and check if the last digit is a zero or not. – GReyes Aug 08 '21 at 04:07
  • Haha nice one. I meant more in base 10. Sorry, i wasn't too specific on that. – user356448 Aug 08 '21 at 04:11
  • 2
    The inductive step of casting out threes is $,a+10b\equiv a+b\pmod{!3}.,$ That's as simple as it gets. Why do you seek "another" method, and how do you define "other", i.e. what is your definition of a test being "other" or "different". – Bill Dubuque Aug 08 '21 at 04:22
  • @BillDubuque trying to see if other methods are possible, that can potentially be used with #s. Yes, sum of digits works, but it seems too simple. By ,,other" i mean any patterns, adding/ multiplying of digits in any form. – user356448 Aug 08 '21 at 04:30
  • Are you familiar with congruences (modular arithmetic)? – Bill Dubuque Aug 08 '21 at 04:57
  • Just learning :).Sorry if my question seems simple/trivial. – user356448 Aug 08 '21 at 05:06
  • Does the prior comment mean you are not familiar with modular arihmetic / congruences? The more context you provide (including your knowlegde level) the more likely you will receive helpful answers. – Bill Dubuque Aug 08 '21 at 05:22
  • You need all the digits to decide whether a number is divisible by $3$, therefore there is no easier rule. A rule needing only the last digits exists if and only if the number of your interest has no prime factors other than $2$ and $5$, for example for $125$, we need only the last three digits. – Peter Aug 08 '21 at 06:12

1 Answers1

1

Let $x$ and $y$ be the count of the digits $\equiv 1 \pmod 3$ (i.e., $1,4,$ or $7$) and $\equiv 2 \pmod 3$ (i.e., $2,5,$ or $8$) respectively. Then, $x-y$ is divisible by $3$ (or $x \equiv y \pmod 3$) if and only if the original number is.

This is the second condition for the divisor $3$ at the Wikipedia article on divisibility rules.

Geoffrey Trang
  • 9,959
  • But that's essentially equivalent to casting out threes, i.e. reduce the digits mod $3$ then add them. The above simply groups together the digits $\equiv 1$ (viz., $1,4,7)$ and the digits $\equiv -1$ (viz. $,2,5 ,8),,$ i.e. it's just a rearrangement of the sum used in casting out threes. – Bill Dubuque Aug 08 '21 at 04:51
  • That's what i meant! @ Geoffrey Tranq. Are there any other for #3? – user356448 Aug 08 '21 at 04:58