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Maybe the title of what im trying to describe isn't right (i'm dutch) and i can't find it anywhere because of that. When i was in elementary school they learned me something called the Nine rule (directly translate from dutch "De negen proef."). Its a simple way to control your sumation, subtraction, multiplication, division. So my question is how is this called in english and is there a mathematical proof for it? example:

13 + 15 = 28 $\rightarrow$ control (1+3) + (1+5) = 4+6 = 10 = (1+0) = 1 == (2+8) = 10 = (1+0) = 1 $$\begin{array}{lcr|c} & 1 & 3 & 4 \\ \text{+} & 1 & 5 & 6 \\ \hline & 2 & 8 & \text{1\1} \\ \end{array}$$

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    $\begingroup$ In America, at least, that is called "casting out nines". en.wikipedia.org/wiki/Casting_out_nines $\endgroup$
    – user247327
    Jan 20, 2016 at 22:49
  • $\begingroup$ Thanks a lot i found the proof already! $\endgroup$
    – user306660
    Jan 20, 2016 at 22:53

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I don't know the english name, but it's quite easy to prove. It simply checking that your computation holds modulo 9 (as any number is equal modulo 9 to the sum of its digits). Here

$13 = 4~mod~9$ (as $13=1\times 9+4$)

$15 = 6~mod~9$ (as $15=1\times 9+6$)

thus

$13+15=6+4=10=1~mod~9$ (and indeed $28=3\times 9+1$)

It should be noted that this obviously does not guarrantee that the computation is correct.

It should be noted that it does not work for division :

15 divided by 3 is 5, and 6 divived by 3 is 2.

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I'm not sure if there is a name for it, but the proof is modular arithmetic, combined with the following observation.

Lemma: Given a positive integer, $n$, the sum of the digits of $n$ has the same remainder upon division by $9$ that $n$ does.

Proof. Let us consider $3$ digit numbers for the sake of exposition. We could work in greater generality, but it would require a little bit more notation.

Because $10=9+1$, we will also have that $100=(9+1)(9+1)=9(9+1+1)+1$, and more generally, $10^n=9k_n+1$ for some integer $k_n$ depending on $n$. Thus, $10^n-1$ is divisible by 9 for every $n$.

Now, write a number whose digits are $n=[x][y][z]=100x+10y+z$. Then $n$ minute the sum of the digits of $n$ is $(100x+10y+z)-(x+y+z)=99x+9y$ is a multiple of $9$.Therefore, the remainder upon dividing $n$ by $9$ is the same as the remainder upon dividing $(x+y+z)$ by $9$.


So what the lemma tells us is that by adding the digits of a number (and adding those digits again if necessary), you will eventually get the remainder upon division by $9$. The last step we need is that, if we add two numbers, their remainders add, and if we multiply two numbers, their remainders multiply.

To see this, let us add $9a+b$ and $9c+d$ to get $9(a+c)+(b+d)$. We see that dividing this sum by $9$, we will get the same remainder as if we divided $(b+d)$ by $9$.

Similarly, if we multiply, $(9a+b)(9c+d)=9(9ac+ad+bc)+bd$, and if we divide this by $9$, we get the same remainder as if we had divided $bd$ by $9$.


To summarize, we have two things going on. We can add digits to find the remainder upon division by $9$, and we know that the remainder of a sum/product is the same as the sum/product of the remainers (assuming we divide by $9$ and take the remainder if we get something larger than $9$).

All this can be generalized. If you search for "modular arithmetic" you will find a lot of resources on this kind of math.

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