enigma123 wrote:
If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal?
(1) a > c
(2) b > d
Any idea what is the concept behind this question to get a answer B?
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 \(b\) (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 to \(2*5^3\). 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 the terminating decimal.
We need reducing in case when we have the 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:
If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal?Question: is \(\frac{2^a*3^b}{2^c*3^d*5^e}\) a terminating decimal? The question basically asks whether we cans reduce 3^d in the denominator so to have only powers of 2 and 5 left, which can be rephrased is b (the power of 3 in the nominator) greater than or equal to d (the power of 3 in the denominator): is b>=d?
(1) a > c. Not sufficient.
(2) b > d. Sufficient.
Answer: B.
Hope it helps.
as you said that both the numerator and denominator should be reduced to the lower term ,,, but as in the question, it was clearly seen that 2^a and 2^c is present in both numerator and denominator so the term is not reduced to its lowest term...
now from statement 1, we can clearly state that 2 will not present in the denominator and we have to have only 2 and 5 in the denominator to say that the term would be terminating. Hence statement 1 looks sufficient to me