Question

Can anyone figure out a trick to solve this?
If p = 216^–1/3 + 243^–2/5 + 256^–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

I can supply the answer if necessary but the only way I know to arrive to it is by a LOT of trial and error : /. It's GRE so I know there must be a trick to it!

Answer 1

\[ p = 216^{–1/3} + 243^{–2/5} + 256^{–1/4}\]
*A little cleaner

Answer 2

216^-1/3 = (6^3)^-1/3\[(216)^{-1/3} = (6^{3})^{-1/3} = 6^{3*-1/3} = 6^{-1}\]
\[(243)^{-2/5} = [(243)^{2}]^{-1/5} = [59049]^{-1/5} = (9^{5})^{-1/5} = 9^{5*-1/5}= 9^{-1}\]
\[(256)^{-1/4} = (4^{4})^{-1/4} = 4^{4*-1/4} = 4^{-1}\]
so the given equation reduces to
p = 216^–1/3 + 243^–2/5 + 256^–1/4 = 6^-1 + 9^-1 + 4^-1

Answer 3

How does one come to this conclusion w/o a calculator tho (considering one cannot use a calc on the GRE)? Is there a property I'm unaware of or is one simply expected to memorize powers?

Answer 4

well u hv to be aware about exponential rules and also the hint is in the denominator of the power...
in base number can be reduced to some number with power = denominator of the outer power as I hv done.....
then u hv to be aware that 2 multiplied to itself so may times will end in 4 or 6 or 8 and so on

once u know these things, you can do the calculations orally and quickly find the final result.....

Answer 5

which one of the following is an integer? (A)p/19 (B) p/36 (C) p (D)19/p (E) 36/p

1 1 1 6 + 4 + 9 19
p = ----- + ------ + ----- = ----------- = ------
6 9 4 36 36
means

36 = 19/p

so 19/p is an integer

Answer 6

So u basically, to start off w/, u looked at the denominators and said...

for (216)^−1/3: what n^3 = 216

and then for (243)^−2/5: what n^5 = 243

and finally for (256)^−1/4: what n^4 = 256

right?

Answer 7

absolutely correct.....
once u hv identified the numbers ie. the n's then rest is all mental/oral....
with a little bit of writing thrown in...☺

Answer 8

got a medal for me ?? .....☺

Answer 9

haha, yes
ok. thank u for your help!