2.) (2 pts) in lowest terms, give the fraction for the repeating decimal \( 0.321321321321 . \mathrm{n}=0.321 \)

To convert the repeating decimal \( 0.321321321... \) into a fraction, we can denote it as \( x \). Thus, \( x = 0.321321321... \). To eliminate the repeating part, we can multiply both sides by \( 1000 \) (since there are three digits in the repeat): \[ 1000x = 321.321321321... \] Now, subtract the original equation from this new one: \[ 1000x - x = 321.321321321... - 0.321321321... \] This simplifies to: \[ 999x = 321 \] Now, divide both sides by 999: \[ x = \frac{321}{999} \] To express this fraction in its lowest terms, we can find the greatest common divisor (GCD) of 321 and 999. The GCD is 3. Therefore, we can divide both the numerator and denominator by 3: \[ x = \frac{321 \div 3}{999 \div 3} = \frac{107}{333} \] So, the repeating decimal \( 0.321321321... \) can be expressed as the fraction \( \frac{107}{333} \) in its lowest terms.