I know of the sum of digits divisible by 3 method, but it seems to not be working for base 5.

How can I check if number in base 5 is divisible by 3 without converting it to base 10 (or 3, for that matter)?

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    $\begingroup$ The sum of digits method is based on the fact that $3$ divides $10-1$. So on base $5$ you could apply this method in order to check divisibility by $2$ or by $4$, but not by $3$. $\endgroup$ – barak manos Jun 6 '16 at 18:57
  • $\begingroup$ @barakmanos Hrm. But isn't there some other way? $\endgroup$ – Akiiino Jun 6 '16 at 18:57
  • $\begingroup$ Dunno. For divisibility by $6$ on base $5$, you could use the same method used for divisibility by $11$ on base $10$. That should cover half of the cases. Not sure about the other half though... $\endgroup$ – barak manos Jun 6 '16 at 18:58

Divisibility rules generally rely on the remainders of the weights of digits having a certain regularity. The standard method for divisibility by $3$ in the decimal system works because the weights of all digits have remainder $1$ modulo $3$. The same is true for $9$. For $11$, things are only slightly more complicated: Since odd digits have remainder $1$ and even digits have remainder $-1$, you need to take the alternating sum of digits to test for divisibility by $11$.

In base $5$, we have the same situation for $3$ as we have for $11$ in base $10$: The remainder of the weights of odd digits is $1$ and that of even digits is $-1$. Thus you can check for divisibility by $3$ by taking the alternating sum of the digits.

More generally, in base $b$ the sum of digits works for the divisors of $b-1$ and the alternating sum of digits works for the divisors of $b+1$.

  • 4
    $\begingroup$ Another way of saying the same, since $5\equiv -1 \pmod{3}$: If we have a number (backwards) $a_0+a_1\cdot 5+a_2\cdot 5^2+a_3\cdot 5^3+\dots +a_n\cdot 5^n$ then that is congruent to $a_0+a_1\cdot (-1)+a_2\cdot (-1)^2+a_3\cdot (-1)^3+\dots +a_n\cdot (-1)^n = a_0-a_1+a_2-a_3+\dots +(-1)^na_n$, so you check divisibility by three by regarding the alternating sum of digits. $\endgroup$ – Jeppe Stig Nielsen Jun 6 '16 at 21:20
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    $\begingroup$ And while were at it: The “last $n$ digits” method works for divisors of $b^n$. $\endgroup$ – Hermann Döppes Jun 6 '16 at 22:01

Add the digits, but multiply the even digits by 2.

This works because $5 \equiv 2 \mod 3$, $5^2 \equiv 1 \mod 3$, etc.

  • 5
    $\begingroup$ Equivalently, add the digits but negate the even-indexed ones, because $5\equiv -1\pmod{3}$. $\endgroup$ – jwodder Jun 6 '16 at 21:54

The $n=(d_m \cdots d_0)_5$ then $n$ is divisible by $3$ iff $d_0-d_1+d_2-d_3+\cdots$ is divisible by $3$.

This follows from $5^k \equiv 1 \bmod 3$ if $k$ is even and $5^k \equiv -1 \bmod 3$ if $k$ is odd.


Hint $ $ Radix notation has Polynomial form $\,n = d_0\! + d_1 5 + d_2 5^2\! +\cdots + d_k 5^k\! = P(5)\,$ so

${\rm mod}\ 3\!:\ \color{#c00}5\equiv \color{#c00}{-1}\,\Rightarrow\ n = P(\color{#c00}5) \equiv P(\color{#c00}{-1}) \equiv d_0 - d_1 + d_2 - \cdots + (-1)^k d_k\, $ by applying the Polynomial Congruence Rule, i.e. $\,a\equiv b\,\Rightarrow\,P(a)\equiv P(b)$.

Remark $\ $ This is essentially the same rule for casting $11$s in decimal notation, i.e. compute the alternating sum of the digits modulo the radix. Clearly the same method works whenever the radix $\equiv -1$ modulo the divisor.

Similar ideas tackle higher degree cases, e.g. casting out $91 = 10^2\!-10+1$ works like this:

$$x^2 \equiv x-1\ \Rightarrow\ x^3 \equiv -1,\,\ x^4 \equiv -x,\,\ x^5 \equiv 1-x,\,\ x^6 = 1 \pmod{x^2 - x + 1}$$

Hence $\ d := d_0 + d_1 x + d_2 x^2 + d_3 x^3 + d_4 x^4 + d_5 x^5 + d_6 x^6$

$\qquad\quad\ \equiv d_0-d_2-d_3+d_5+d_6 + (d_1+d_2-d_4-d_5)\, x\,\ \pmod{x^2 - x + 1}$

E.g. for $\, x=10\ $ the modulus is $\,x^2-x+1 = 91\,$ so we have

$$ d=6543210 \equiv 0\! -\!2\! \!-3\! +\!5\!+\!6 + (1\! +\!2\! -\!4\! -\!5) 10 \equiv 6-60 \equiv \color{#0a0}{37}\! \pmod{\!91}$$

And indeed $\ 6543210 = \color{#0a0}{37} + 91 \cdot 71903.$


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