The question is how do I prove any number is not divisible by $3$. So for example, how do I prove $481$ is not divisible by $3$?
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4Add the numbers:$$4+8+1=13\ne\text{divisible by three}$$ – Simply Beautiful Art Feb 18 '17 at 22:34
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Also, when adding the digits, you can ignore the digits that are multiples of 3 (0, 3, 6, and 9). – Wood Feb 18 '17 at 22:42
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Possible duplicate of Prove "casting out nines" of an integer is equivalent to that integer modulo 9 – hardmath Feb 19 '17 at 16:06
5 Answers
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Since $10 \equiv 1 \pmod 9$, $\sum a_i \equiv \sum a_i 10^i \pmod 9$. Hence $481 \equiv 4 + 8 + 1 = 13 \equiv 4 \pmod 9$ hence is not divisible by 3.
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jay_Hathaway has the only answer that has a proof. The others just show the test he proves. – stretch Feb 18 '17 at 22:48
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@stretch: Well, to be pedantic, this shows that it is not divisible by 9, which is true, but the question was if it was divisible by 3. The same idea applies, but with 3 rather than 9. – copper.hat Feb 19 '17 at 00:05
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As "Simply Beautiful Art' notes you can tell whether or not a number is divisible by three by whether or not the digits within it are divisible by three. Because the digits of 481 sum to 13 which is clearly not divisible by three, 481 is not divisible by three.
Another way to prove this would be to find the prime factorization of 481, that is divide this number into it's prime factors. The prime factorization of 481 is 13x37. Because we don't see a 3 in the prime factorization we can be sure that this number is not divisible by three.
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Since $10 \pmod 3 = 1$, we have $(\sum_k d_k 10^k) \pmod 3 = (\sum_k d_k) \pmod 3$.
Hence a number is divisible by $3$ iff the sum of the digits is divisible by 3.
Hence $481 \pmod 3 = 13 \pmod 3 = 4 \pmod 3 = 1 \pmod 3$ hence it is not divisible by $3$.
copper.hat
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A number is divisible by three if and only if its digits add up to a number divisible by three. So since, as Simply Beautiful Art noted, 481's digits add up to 13, which is not divisible by 3, 481 itself is also not divisible by three.
The Count
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$$A=\overline{...a_3a_2a_1a_0}\\A=a_0+10a_1+100a_2+10^3a_3+...=\\ A=a_0+(1+9)a_1+(1+99)a_2+(1+999)a_3+...=\\ (a_0+a_1+a_2+....)+9(a_1+11a_2+111a_3+...)=\\ (a_0+a_1+a_2+....)+9k$$ Now if $3|A$ we have $$3|A=(a_0+a_1+a_2+....)+9(a_1+11a_2+111a_3+...)\\ 3|(a_0+a_1+a_2+....)+9k\\ 3|(a_0+a_1+a_2+....)\\$$ show that $481\neq3k$ proof by cotradittion
first suppose $481=3k$ $$481=1+8\times 10 +4\times 100=(1+8+4)+9(1\times 8+4\times 11)=3k\\(1+8+4)+9q=3k\\(1+8+4)=3k-9q=3(k-3q)=3q' \\\to 13=3q' \\$$since it is not correct
Khosrotash
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