Bunuel wrote:
Official Solution:
THEORY:
Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only if \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).
Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.
For example \(\frac{x}{2^n5^m}\), (where \(x\), \(n\) and \(m\) are integers) will always be a terminating decimal.
(We need reducing in case we have a prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.)
BACK TO THE ORIGINAL QUESTION:
(1) \(s = 225\). Hence \(\frac{r}{s^2}=\frac{r}{225^2}=\frac{r}{9^2*5^4}\). We don't know whether \(9^2\) can be reduced, so we cannot say whether this fraction will be a terminating decimal.
(2) \(r = 81\). Not sufficient on its own.
(1)+(2) \(\frac{r}{s^2}=\frac{9^2}{9^2*5^4}=\frac{1}{5^4}\), since denominator has only 5 as prime, then this fraction is a terminating decimal.
Answer: C
Bunuel can you explain in more detail please ? Does this mean as long as reduced form of the fraction has positive power of either 2 or 5 it will be a terminating fraction ?
If a fraction no matter whether it's reduced or not has only positive integer powers of 2's or 5's in the denominator the fraction will be terminating.
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