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Question id: 137. This question is similar to one that appeared in a GCSE Higher paper (specimen) for 2017. The use of a calculator is allowed.

Proof

(a) Prove that the recurring decimal \(0.\dot2 \dot1\) has the value \(\frac{7}{33}\)

(b) The value of \(x\) is given as:

$$x=\frac{1}{5^{120}\times2^{123}}$$

Show that, when \(x\) is written as a terminating decimal, there are 120 zeros after the decimal point.

(c) The reciprocal of any prime number \(p\) (where \(p\) is neither 2 nor 5) when written as a decimal, is always a recurring decimal.

A theorem in mathematics states:

The period of a recurring decimal is the least value of \(n\) for which \(p\) is a factor of \(10^n – 1\)

Marilou tests this theorem for the reciprocal of eleven.

She uses her calculator to show that 11 is a factor of \(10^2 – 1\) then makes this statement:

"The period of the recurring decimal is 2 because 11 is a factor of \(10^2-1\). This shows the theorem to be true in this case."

Explain why Marilou's statement is incomplete.

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