Number Theory

Rational Numbers

Rational Numbers: Level 3 Challenges


True or False?

0.9999=10.9999 \ldots = 1

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

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

A=0.19+0.199,B=0.19×0.199{A = 0.\overline{19} + 0.\overline{199}, \quad B = 0.\overline{19} \times 0.\overline{199}}

Recall that 0.19,0.\overline{19}, for example, stands for the repeating decimal 0.19191919...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 0.190.\overline{19} is 2.

Find the sum of the periods of AA and BB.

Let aka_k represent the repeating decimal 0.133k0.\overline{133}_k for k4k \geq 4. The product a4a5a99a_4 a_5 \cdots a_{99} can be expressed as mn!\frac{m}{n!} where m,nm, n are positive integers and nn is as small as possible. mn\frac{m}{n} can be expressed as pq \frac{p}{q} where p,qp, q are coprime integers. What is p+qp+q?

Note: 0.133k0.\overline{133}_k refers to the repeating decimal 0.133133133 0.133133133\ldots evalauted in base kk.

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

S=19+199+1999++110n1+ \large S=\frac { 1 }{ 9 } +\frac { 1 }{ 99 } +\frac { 1 }{ 999 } +\ldots + \frac { 1 }{ { 10 }^{ n }-1 } + \ldots


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