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1997 Paper 1 Q3
D: 1484.0 B: 1501.4
sequences recurrence relation exponentials and logarithms powers of 3 modular arithmetic estimation decimal approximation

Let \(a_{1}=3\), \(a_{n+1}=a_{n}^{3}\) for \(n\geqslant 1\). (Thus \(a_{2}=3^{3}\), \(a_{3}=(3^{3})^{3}\) and so on.)

  1. What digit appears in the unit place of \(a_{7}\)?
  2. Show that \(a_{7}\geqslant 10^{100}\).
  3. What is \(\dfrac{a_{7}+1}{2a_{7}}\) correct to two places of decimals? Justify your answer.

Solution:

  1. Notice that \(a_n = 3^{3^{n-1}}\) in particular, \(a_7 = 3^{3^6}\). Using Fermat's little theorem, we can see that \(3^4 \equiv 1 \pmod{5}\) and so we need to figure out \(3^6 \pmod{4}\), which is clearly \(1\). Therefore \(3^{3^6} \equiv 3^{4k+1} \equiv 3 \pmod{5}\). Therefore the units digit is \(3\).
  2. Notice that \(3^5 > 100\) and \(3^3 > 10\). Therefore \begin{align*} a_7 &= 3^{3^6} \\ &= (3^3)^{3^5} \\ &> 10^{3^5} \\ &> 10^{100} \end{align*}
  3. \begin{align*} \frac{a_7+1}{2a_7} &= \frac12 + \frac1{2a_7} \\ &= 0.5 + 0.\underbrace{000\cdots}_{\text{at least }99\text{ zeros}} \\ &= 0.50 \end{align*} Since \(a_7 > 10^{100}, \, \frac{1}{2a_7} < 10^{-100}\)

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