# DFA that accepts decimal representations of a natural number divisible by 43

First, I have tried to build a DFA over the alphabet $\sum = \{0,\dots, 9\}$ that accepts all decimal representations of natural numbers divisible by 3, which is quite easy because of the digit sum. For this I choose the states $Q = \mathbb{Z}/3\mathbb{Z}\cup\{q_0\}$ ($q_0$ to avoid the empty word), start state $q_0$, accept states $\{[0]_3\}$ and $\delta(q, w) =\begin{cases} [w]_3 &\mbox{if } q = q_0 \\ [q + w]_3 & \mbox{else } \end{cases}$

Of course, it doesn't work that way for natural numbers divisible by 43. For 43 itself, I would end in $[7]_{43}$, which wouldn't be an accepting state. Is there any way I can add something there or do you have other suggestions on how to do this? Thanks.

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– Paresh Jan 20 at 9:15

4
– A.Schulz Jan 20 at 10:58

## 1 Answer

If the string read has a certain decimal value, then reading the next digit changes that value: multiply by $10$ and add that digit. The DFA keeps track of that value modulo $43$. Thus, for $q\in \{0,1,\dots,42\}$ and $a\in \{0,1,\dots,9\}$ you do

$\qquad \delta(q,a) = (10\cdot q + a) \bmod 43$.

Note that the DFA does not actually perform the computation; the transition is hard-coded.

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