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 non-negative integers and p=2^a*3^b and q=2^c*3^d*5^e, is p/q a terminating decimal?

(1) a > c

(2) b > d
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Can you explain why A is the answer?

I guess A only tells us that 2 wont be there in the denominator but does not tell us anything about 3, and now if 3 will be there in the denominator it will be non terminating decimal but if 3 wont be there then it will be a terminating decimal and hence its not sufficient

on the other hand st 2 clearly tells that 3 wont be there in the denominator and hence its sufficient.

Correct me if I am missing some thing here ..!

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*5^2. 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.

Hope it helps.
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1. Not sufficient
As 3 will be there in denominator or not.

2.sufficient
As we know 3 is not in the denominator and denominator has a 5.

Answer is B.

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Hi,

p = 2^a3^b and q = 2^c3^d5^e
p/q = 2^(a-c)3^(b-d)5^(-e)

division by 2 or 5 (raised to any power will result to terminating decimal)

only check is on b-d

so, using (2)
b-d > 0,
thus 3 is not in denominator, hence p/q is terminating decimal.

Answer is (B).

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My answer is B. What's the OA?

The only determinant if p/q is a terminating decimal is 3^b/3^d since a fraction with the format a/b is terminal if b is a power of 2 or 5 or both.
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OA is given in the initial post and it's B.

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 non-negative integers and p=2^a*3^b and q=2^c*3^d*5^e, is p/q a terminating decimal?

(1) a > c
(2) b > d

Check this: any-decimal-that-has-only-a-finite-number-of-nonzero-digits-90504.html#p689656

So, according to that \frac{p}{q}=\frac{2^a*3^b}{2^c*3^d*5^e} will be terminating decimal if 3^d in the denominator can be reduced, which will happen when the power of 3 in the numerator is more than or equal to the power of 3 in the denominator, so when b\geq{d}.

As we can see (1) is completely irrelevant to answer whether b\geq{d}, while (2) directly answers the question by stating that b>d.

Answer: B.

Hope it's clear.
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Not sure I understand what you mean above. What difference does it make whether 2^a is reduced or not? Or whether a-c is positive?

Please read this: any-decimal-that-has-only-a-finite-number-of-nonzero-digits-90504.html#p689656 and this: any-decimal-that-has-only-a-finite-number-of-nonzero-digits-90504.html#p1096496 for theory.

Solution: any-decimal-that-has-only-a-finite-number-of-nonzero-digits-90504.html#p1096500
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p/q = (2^a * 3^b) / (2^c * 3^d * 5^e)
= [2^(a-c) * 3(b-d)] / 5^e

A/B will be terminating(T) only if 1/B is T.
B = 1/5^e will be always T , in the same way as 1/3^e will always be non- terminating(NT).
Product of a NT with a T will always be NT.

In the numerator of p/q, powers of 2 and 3 can be +ve or -ve. Power of 2 doesn't have any affect on the T behaviour of p/q, but if power of 3 is -ve, it will go down to denominator and 1/3^x is always NT and make p/q NT.

stmt 1: a > c means power of 2 will be positive - INSUFFICIENT.
stmt 2: b > d means power of 3 will be positive and 3^(b-d) will remain in the numerator. Thus, p/q will be T. SUFFICIENT

Answer is B
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