Number Theory

Rational Numbers

Rational Numbers: Level 3 Challenges


True or False?

\[0.9999 \ldots = 1\]

Note: The "\(\ldots\)" indicates that there are infinitely many 9's.

Find the number of rational numbers \(r\, (0<r<1)\) such that when \(r\) is written as a fraction in lowest terms, the numerator and denominator sum to 1000.

\[{A = 0.\overline{19} + 0.\overline{199}, \quad B = 0.\overline{19} \times 0.\overline{199}}\]

Recall that \(0.\overline{19},\) for example, stands for the repeating decimal \(0.19191919...\) and that the period of a repeating decimal is the number of digits in the repeating part. In this case, the period of \(0.\overline{19}\) is 2.

Find the sum of the periods of \(A\) and \(B\).

Let \(a_k\) represent the repeating decimal \(0.\overline{133}_k\) for \(k \geq 4\). The product \(a_4 a_5 \cdots a_{99}\) can be expressed as \(\frac{m}{n!}\) where \(m, n\) are positive integers and \(n\) is as small as possible. \(\frac{m}{n}\) can be expressed as \( \frac{p}{q}\) where \(p, q\) are coprime integers. What is \(p+q\)?

Note: \(0.\overline{133}_k\) refers to the repeating decimal \( 0.133133133\ldots \) evalauted in base \(k\).

If the following infinite series \(S\) is evaluated as a decimal,what is the 37th digit to the right of the decimal place?

\[ \large S=\frac { 1 }{ 9 } +\frac { 1 }{ 99 } +\frac { 1 }{ 999 } +\ldots + \frac { 1 }{ { 10 }^{ n }-1 } + \ldots \]


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