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If p = 216^(–1/3) + 243^(–2/5) + 256^(–1/4), then which one of the

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If p = 216^(–1/3) + 243^(–2/5) + 256^(–1/4), then which one of the  [#permalink]

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New post 10 Apr 2019, 01:15
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A
B
C
D
E

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  65% (hard)

Question Stats:

38% (02:07) correct 63% (02:01) wrong based on 16 sessions

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Re: If p = 216^(–1/3) + 243^(–2/5) + 256^(–1/4), then which one of the  [#permalink]

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New post 10 Apr 2019, 02:05
P= 1/6+ 1/9 + 1/4
=19/36
Therefore 19/p=36(which is an integer)
IMO D

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Re: If p = 216^(–1/3) + 243^(–2/5) + 256^(–1/4), then which one of the  [#permalink]

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New post 10 Apr 2019, 02:56
Bunuel wrote:
If \(p = 216^{–\frac{1}{3}} + 243^{–\frac{2}{5}} + 256^{–\frac{1}{4}}\), then which one of the following is an integer?

(A) p/19
(B) p/36
(C) p
(D) 19/p
(E) 36/p


solve \(p = 216^{–\frac{1}{3}} + 243^{–\frac{2}{5}} + 256^{–\frac{1}{4}}\)
we get = 19/36=p
for given options 19/p will be an integer
IMO D
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Re: If p = 216^(–1/3) + 243^(–2/5) + 256^(–1/4), then which one of the   [#permalink] 10 Apr 2019, 02:56
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