0

So the given number is $1010908899$ and I showed that it is divisible by $11$ since $9-9+8-8+0-9+0-1+0-1=-11$ which is divisible by $11$ hence the number also. But how to check if it is also divisible by $7$ and $13$?

Kushal Bhuyan
  • 7,220
  • See this answer. If I'm not mistaken, then you do it for each one of these factors in the same way you did it for $11$, but in groups of $3$ digits instead of $1$ (so in the example above, you check $101-090+889-9\color\red{00}$). – barak manos Nov 10 '15 at 14:27
  • If you don't need to do it by hand then factor the number using an online tool. $1010908899 = 3^2 \cdot 7 \cdot 11^2 \cdot 13 \cdot 101^2 $ – John Alexiou Nov 10 '15 at 14:28
  • ... or just use long division. – Michael Biro Nov 10 '15 at 14:28
  • I want to do it in number theory way, using general results. – Kushal Bhuyan Nov 10 '15 at 14:29
  • I have a divisibility test for 7 . If you want to find out whether a number is divisible by 7 or not then just square the last digit of the number and subtract it from the remaining number . If the number is divisible by 7 then the original number is also divisible . so here last digits is 9 so $9^2$=81 so remaining number - the square=101090889-81=101090808 which is divisible by 7 ( by calculator) thus the number 1010908899 is too divisible by 7. Hope this helps you. – Archis Welankar Nov 10 '15 at 15:08

3 Answers3

5

Since $7$, $11$, and $13$ are prime numbers, a number is divisible by $7$, $11$, and $13$ if and only if it is divisible by $7 \cdot 11 \cdot 13 = 1001$.

N. F. Taussig
  • 79,074
3

The test I've seen for divisibility by $7$ is to chop off the one's digit, multiply by 2, and subtract from the number that was left over. Keep going until you get an answer that is obviously (or obviously not) a multiple of $7$. Thus $$\begin{array}{lll} 1010908899 &\to &101090889 - 18 = 101090871 \\ 101090871 & \to & 10109087 - 2 = 10109085 \\ 10109085 & \to & 1010908 - 10 = 1010898 \\ 1010898 & \to & 101089 - 16 = 101073 \\ 101073 &\to & 10107 - 6 = 10101 \\ 10101 &\to & 1010 - 2 = 1008 \\ 1008 &\to & 100 - 16 = 84 \\ 84 &\to & 8 - 8 = 0 \end{array} $$

Since $0$ is a multiple of $7$, so was $1010908899$. This isn't really a practical way to do it, but it is kind of interesting.

A similar test for divisibility by $13$ is to chop off the one's digit, multiply by $4$, then add to the number that was left over. In the case of your number, $$\begin{array}{lll} 1010908899 &\to &101090889 +36 = 101090925 \\ 101090925 & \to & 10109092 + 20 = 10109112 \\ 10109112 & \to & 1010911 +8 = 1010919 \\ 1010919 & \to & 101091 + 36 = 101127 \\ 101127 &\to & 10112 +28 = 10140 \\ 10140 &\to & 1014 + 0 = 1014 \\ 1014 &\to & 101 + 16 = 117 \\ 117 &\to & 11 +28 = 39 \\ 39 &\to & 3 + 36 = 39 \end{array} $$ which repeats indefinitely, but $39$ is a multiple of $13$, and thus so is $1010908899$.

Umberto P.
  • 54,204
  • You don't need to go quite so far, because $999999$ is divisible by $7$, so you can first take the sum of ever group of six digits: $908899+1010$. You can then apply the double-and-subtract. – Thomas Andrews Nov 10 '15 at 14:42
  • I'm upping one because I think this is a simple divisibility rule that should be better known - and definitely more often taught in schools. Instead, we're often taught there's no easy divisibility rule for $7$. By the way, the reason this works is that we're transforming $100x + 10y + z$ into $10x + y - 2z$. Modulo $7$, the former is $2x + 3y + z$ and the latter is $3x + y - 2z$. If the latter is a multiple of $7$, $y = 2z - 3x \pmod 7$. Substituting that back into the former gives $10x -3x = 7x = 0 \pmod 7$. So divisibility of the "reduced" form implies divisibility of the original. – Deepak Nov 10 '15 at 14:58
  • I only covered the rule for $7$, but one could see why the rule for $13$ works the way it does using similar methods. – Deepak Nov 10 '15 at 15:01
  • @Deepak it is somewhat simpler than that: $10x + y \equiv 0 (\bmod 7)$ implies $30x + 3y \equiv 0 (\bmod 7)$ implies $-20 x - 2y \equiv 0 (\bmod 7)$ implies $x - 2y \equiv 0 (\bmod 7)$ and vice versa, since 21 is a multiple of 7. – Umberto P. Nov 10 '15 at 15:09
2

General results:

$$ \begin{array}{|c|c|c|c|c|} \hline n & 10^n & 10^n \bmod 7 & 10^n \bmod 11 & 10^n \bmod 13 \\ \hline 0 & 1 & 1 & 1 & 1 \\ 1 & 10 & 3 & -1 & -3 \\ 2 & 10^2 & 2 & 1 & -4 \\ 3 & 10^3 & -1 & -1 & -1 \\ 4 & 10^4 & -3 & 1 & 3 \\ 5 & 10^5 & -2 & -1 & 4 \\ \hline 6 & 10^6 & 1 & 1 & 1 \\ 7 & 10^7 & 3 & -1 & -3 \\ \cdots & \cdots & \cdots & \cdots & \cdots \end{array} $$

So, the number $$...d_7d_6d_5d_4d_3d_2d_1d_0$$ is divisible by $7$, iff $$(1,3,2,-1,-3,-2,\; 1,3,...)\cdot(d_0,d_1,d_2,d_3,d_4,d_5,d_6,d_7,...)\equiv 0(\bmod 7),$$ is divisible by $11$, iff $$(1,-1,1,-1,1,-1,\; 1,-1,...)\cdot(d_0,d_1,d_2,d_3,d_4,d_5,d_6,d_7,...)\equiv 0(\bmod 11),$$ is divisible by $13$, iff $$(1,-3,-4,-1,3,4,\; 1,-3,...)\cdot(d_0,d_1,d_2,d_3,d_4,d_5,d_6,d_7,...) \equiv 0(\bmod 11).$$
Here we use dot product: $(u_0,u_1,u_2,...)\cdot (v_0,v_1,v_2,...) = u_0v_0+u_1v_1+u_2v_2+...$.

So, on divisibility by $7$:
$1010908899 \equiv (1,3,2,-1,-3,-2,1,3,2,-1)\cdot (9,9,8,8,0,9,0,1,0,1)=28\equiv 0 (\bmod 7)$,
on divisibility by $11$: $1010908899 \equiv (1,-1,1,-1,1,-1,1,-1,1,-1)\cdot (9,9,8,8,0,9,0,1,0,1)=-11\equiv 0 (\bmod 11)$,
on divisibility by $13$: $1010908899 \equiv (1,-3,-4,-1,3,4,1,-3,-4,-1)\cdot (9,9,8,8,0,9,0,1,0,1)=-26\equiv 0 (\bmod 13)$.

Oleg567
  • 17,740