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Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals. If a, b, c, d and e are integers and p = 2^a*3^b and q = 2^c*3^d*5^e, is p/q a terminating decimal?

Its B. (1) a > c is not suff because b and d are unknown. NSF (2) b > d is suff because 2^(a-c) and 5^e is always terminating decimal isrrespective of the values of a, c and e. Suff.
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\(p/q = ((2^a)*(3^b))/((2^c)*(2^d)*(5^e)) = ((2^(a-c))*(3^(b-d)))/(5^e)\) If \(e>0\), then we have some multiple of 5 in the denominator, and that's ok, it will terminate. Likewise if \(e<=0\) then we have a bunch of 5's up top. If we have a bunch of 2's in the denominator, then it's irrelevant too. The only problem is if we have 3's in the denominator, if we do, it's repeating, if not, it's not repeating. How do we ensure there are no 3's down below? Make sure the exponent on the 3 is positive, or \(b-d>0\) or \(b>d\).

Looking at the answer choices, the answer is clearly \(B\).
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I understand the explanations above. However, I'm also interested in other scenarios. For example, let's say you break a fraction's numerator and denominator down to their prime factorizations and then reduce the original fraction to yield the reduced fraction. Will the presence of any prime numbers, other than 2 and 5, in the denominator of the reduced fraction yield a repeating (i.e. non-terminating) decimal?

I understand the explanations above. However, I'm also interested in other scenarios. For example, let's say you break a fraction's numerator and denominator down to their prime factorizations and then reduce the original fraction to yield the reduced fraction. Will the presence of any prime numbers, other than 2 and 5, in the denominator of the reduced fraction yield a repeating (i.e. non-terminating) decimal?

Yes, that's precisely the rule: if you have reduced your fraction (that step is crucially important), and there's a prime factor of the denominator different from 2 or 5, you'll get a repeating decimal. If 2 and/or 5 are the only prime factors of your denominator, the decimal terminates.

It's easier to see why this should be true by thinking of the opposite problem: how you would write a terminating decimal as a fraction. It has to be possible to write any terminating decimal as a fraction with some power of 10 in the denominator - that is, with no primes besides 2 or 5 in the denominator. For example, if you start with 0.056, you can write that as the fraction 56/1000. Now sometimes when you reduce your fraction, some or all of the 2s or 5s might cancel (as they do here: 56/1000 = 7/125 = 7/(5^3) ), but no new prime besides 2 or 5 could ever show up. Conversely, if you start with a reduced fraction and there's some prime besides 2 or 5 in the denominator, there's not going to be any way to rewrite that fraction (by multiplying the top and bottom by the same number) to ever get a power of 10 in the denominator, because that other prime will always be there.
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Is another reason A is wrong because 2^c could be 0?

2^c cannot be zero for any value of c.

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?

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