# Practice Paper 4 Question 20

The *Csatian* language has the following two rules: *(i) \(B\) is a word,* and *(ii) if \(x\) is a word then \(BBBxxx\) is a word*.

Give an expression for the number of \(B\)s in a valid word.

Suggest a third rule

*(iii)*such that all multiples of 3 \(B\)s are valid words, and the only other valid word is a single \(B.\)Replace rule

*(iii)*with the following:*(iii) if \(BBBBx\) is a word then \(x\) is a word*. Find all pairs \((p,q)\) such that, for*all*integers \(n\ge0,\) the expression \(pn+q\) gives the lengths of all valid words.

## Related topics

## Warm-up Questions

- Find a closed form solution to the recurrence defined by \(a_0=5, a_{n+1}=2a_n+1.\)
- The
*Climb*language has two rules:*(i)*it contains the word \(AAAAAAAA,\) and*(ii)*if \(xx\) is a word, then \(x\) is a word. Write down all the words in the language. - Give the possible values of \(b\) in \(3 \cdot 2^i \equiv b \mod 12\) for \(i \in \{0,1,2,3, \ldots\}.\)

## Hints

- Hint 1(Part (a)) Find and solve a recurrence relation for the length of possible words.
- Hint 2(Part (b)) What is the characteristic of the valid words described by rule
*(ii)*? - Hint 3(Part (b)) ... in terms of their divisibility?
- Hint 4(Part (b)) Rules
*(i)*and*(ii)*only cover the case of a single \(B\) and*some*multiples of 3 \(B\)s. How could rule*(iii)*cover*all*multiples of 3 \(B\)s, given that some multiples of 3 \(B\)s are valid words? - Hint 5(Part (b)) Could you modify rule
*(iii)*given in part (c) to do so?*Note:*Sometimes, reading the next part of the question helps. - Hint 6(Part (c)) From the expression in part (a), find a new expression for the length of the valid words described by the new rule
*(iii)*. - Hint 7(Part (c)) If you subtract \(4k\) from the words described by rule
*(ii)*, do you get all the naturals? - Hint 8(Part (c)) You should notice that by subtracting \(4k\) from the words described by rule
*(ii)*, you are not guaranteed to get all naturals (why?). Which of those naturals are not possible? Try expressing them in a more general form. - Hint 9(Part (c)) Prove that the new expression you get earlier could not be equal to those naturals.
*Hint:*It is sufficient to show that the expression is not equal to the base case of the naturals (why?).

## Solution

We use the recurrence relation \(l_i=3+3l_{i-1},\) which unwinds to give us:\[ \begin{align} l_i&= 3+ 3(3+3l_{i-2}) \\ &=3+3^2+3^2 \cdot l_{i-2} \\ &=\sum_{k=1}^{i}3^k + 3^{i}\cdot l_0 \\ &=\frac{3(3^{i}-1)}{2}+3^{i} \\ &=\frac{5\cdot3^i-3}{2}. \end{align}\]

Factorising \(3\) from \(l_i\) gives \(\frac{5\cdot3^{i-1}-1}{2},\) whose numerator is divisible by \(2,\) which means \(l_i\) is a multiple of \(3.\) Thus, if we want to obtain

*all*multiples of \(3\) as valid words, we can add a rule which subtracts \(3\) \(B\)s from any valid word:*"if \(BBBx\) is a word then \(x\) is a word"*.You can write \(l'_i=\frac{5\cdot3^i-3}{2}-4n,\) where \(i,n\) are the number of applications of rule

*(ii)*and*(iii)*respectively. But the question asks you to simplify this. From (a), \(l_i\) is a multiple of \(3,\) but it does not give*all*multiples of \(3.\) Hence by subtracting \(4n\) from it, you are not guaranteed to get all naturals, so it is not obvious which of the \(4k,4k+1,\)\(4k+2,4k+3\) forms you can generate. The proof is in showing that \(4k\) and \(4k+3\) are not possible. One approach could be:

- Show that there do not exist any positive integers \(i,n\) such that \(\frac{5\cdot3^i-3}{2}-4n=4:\)

Rearranging the given equation yields \(5 \cdot 3^i - 8n = (2 \cdot 4) + 3 = 11,\) i.e. we require that \(5 \cdot 3^i \equiv 11 \mod 8\). But for \(i = 0,1,2,3,4\) we get \(3^i \equiv 1, 3, 1, 3.\) This pattern repeats because:\[ 3^i \equiv 1 \mod 8 \implies 3^{i+1} \equiv 1 \cdot 3 \equiv3 \mod 8 \\ 3^i \equiv 3 \mod 8 \implies 3^{i+1} \equiv 9 \equiv 1\mod 8 \]

Hence, \(3^i \in \{1,3\}\) when taken \(\text{mod }8\). Following that, \(5 \cdot 3^i\) can only be \(5 \cdot 1 \equiv 5\) or \(5 \cdot 3 \equiv 15 \equiv 7\). Hence the equation can never be satisfied.

- Show that there do not exist any positive integers \(i,n\) such that \(\frac{5\cdot3^i-3}{2}-4n=3:\)

Rearranging similarly, the problem becomes \(5\cdot 3^i \equiv 9 \mod 8\). Using a similar argument as above, we get \(5 \cdot 3^i\) is either \(5\) or \(7\) in \(\text{mod }8,\) so the equation can never be satisfied.

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