×
Number Theory

# Rational Numbers: Level 3 Challenges

True or False?

$\large 0.9999 \ldots = 1$


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?

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

×