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mgmat 5

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by resilient » Sun May 25, 2008 8:02 am
If a, b, c, d and e are integers and p = (2^a)(3^b) and q = (2^c)(3^d)95^e), is p/q a terminating decimal?





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




I believe I am missing a terminating decimal rule of some sort. Can anyone tell me where to go to learn more about this? I dont see how choice can make the answer correct.

qa is

b
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Source: — Data Sufficiency |
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by ksh » Sun May 25, 2008 8:34 am
If we take p/q
we get, p/q=2^(a-c)*3^(b-d)*0.2^-e
or p/q=2^(a-c-e)*10^-1*3^(b-d)

1. when a>c, we dont know whether 3 will have positive or negative exponent. in case of negative it will result in non terminating decimal e.g. 2/3=0.6666666.. and so on. NS

2. when b>d it means 3 has positive exponent. even if 2 has negative exponent the result be terminating decimal.

hence b alone is suff.
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mgmat 5

by resilient » Sun May 25, 2008 8:38 am
Nice answer thanks. DO you recommend a way to do this with picking numbers to solidify the answer choices?
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by netigen » Sun May 25, 2008 11:23 am
Terminating decimal rule: the only prime numbers in the denominator that can lead to terminating decimal are 2 and 5

hence, to solve this question we need to make sure that the only factors of the denominator are 2 and 5. To get to this we need to get rid of 3 in the denominator.

hence, b is sufficient.
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hmm

by resilient » Sun May 25, 2008 6:20 pm
no for some reason I lost only that in the transer. Question stands corrected at: Is 3^x<500
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