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For this exercise, I need to calculate all possible numbers that satisfy the eleven test. The eleven test is a generalisation of the following:

Let $a_1, a_2, a_3$ be numbers. Then it should hold that $a_1 + 2a_2 + 3a_3 (\text{mod } 11)=0$

I tried fixing one number and then going on, but that did not work for me:

Let $k$ be a number and let $k:=3a_3$. Therefore, $a_1+2a_2 (\text{mod } 11)=11-k$. But then what? I can fix $l := 2a_2$, then $a_1=11-k-l$.

P..
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    I don’t understand the question. What is the 11 test? Or an 11 test? How is there a test? – Lubin Sep 02 '13 at 15:38
  • Are you trying to prove that, for all $a_1, a_2, a_3 \in \Bbb{Z}$, then $a_1+2a_2+3a_3 \equiv 0 \pmod{11}$? If so, take $a_1 = a_2 = a_3 = 1$; this is a counterexample. Or, are you trying to determine what other conditions exist to for the test to hold? – apnorton Sep 02 '13 at 16:10

2 Answers2

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If you want to achieve $$\sum_{k=1}^rka_k=0\mod 11$$ choose $a_2, a_3 \dots a_r$ arbitrarily and let $$\sum_{k=2}^{r}ka_k=a\mod 11$$

Then $a_1=11-a \mod 11$.

Mark Bennet
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Well, I unfortunately don’t read Dutch, but the elfproef seems to be a rough check that a Dutch account number is valid. You take your nine-digit account number, $d_9d_8d_7\dots d_2d_1$ and calculate $\sum_jjd_j$ modulo $11$. If the result is nonzero, your supposed account number wasn’t valid. So you might as well let the first eight digits $d_9,d_8,\dots d_2$ be arbitrary, calculate the modulo-11 sum above but omitting the $d_1$-term, to get a number $S$ with $0\le S\le10$. Now you want your $d_1$ to be the negative of this, modulo $11$, in other words zero if $S=0$, otherwise $11-S$, and you’re out of luck if $S=1$. So this special case has to be avoided, but I’ll leave that for you to figure out.

Lubin
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