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For integers x, y, and z, if (3^x) (4^y) (5^z) = 3,276,800,0

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For integers x, y, and z, if (3^x) (4^y) (5^z) = 3,276,800,0 [#permalink]

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New post 16 Mar 2013, 10:01
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A
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Question Stats:

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For integers x, y, and z, if (3^x) (4^y) (5^z) = 3,276,800,000 and x + y + z = 15, what is the value of xy/z?

(A) undefined
(B) 0
(C) 3
(D) 5
(E) 15

http://www.veritasprep.com/blog/2010/09 ... -opponent/
[Reveal] Spoiler: OA

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Last edited by Bunuel on 17 Mar 2013, 01:05, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Your Opponent is the exponent [#permalink]

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New post 16 Mar 2013, 10:27
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3,276,800,000 = 32,768 x 100,000 = 2^15 x 100,000 = 2^15 x 5^5 x 2^5 = 2^20 x 5^5

Therefore x =0, y =10, z = 5

xy/z = 0

The answer is B. Although this detailed explanation has been provided, you can straightaway reach the answer by realizing that x=0 since the large number given does not have 3 as a factor, and so xy/z has to be 0.
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Kudos [?]: 74 [0], given: 181

Re: Your Opponent is the exponent [#permalink]

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New post 16 Mar 2013, 11:22
GyanOne wrote:
3,276,800,000 = 32,768 x 100,000 = 2^15 x 100,000 = 2^15 x 5^5 x 2^5 = 2^20 x 5^5

Therefore x =0, y =10, z = 5

xy/z = 0

The answer is B. Although this detailed explanation has been provided, you can straightaway reach the answer by realizing that x=0 since the large number given does not have 3 as a factor, and so xy/z has to be 0.


Thanks :-D but how did you get that the big number = 2^15? and why the absence of 3 from the factors would lead the fraction to be = zero?
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Re: Your Opponent is the exponent [#permalink]

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If we know that 3,276,800,000 is not divisible by 3 (which by the way you can quickly test by adding the digits of the number together and seeing whether they are divisble by 3) , this implies that when it is expressed in the form (3^x) (4^y) (5^z) the x=0. Hence xy/z =0
Concerning the 2^15, either you know the powers of 2 or you don't. This one is indeed very high so it seems unlikely to know it by heart. But if you divide 32768 by 8 for example you land on 4096 (which is 2^12) so you eventually figure out that it's 2^15. But again, this was not necessary in order to solve the question.
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Re: For integers x, y, and z, if (3^x) (4^y) (5^z) = 3,276,800,0 [#permalink]

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New post 07 Sep 2013, 13:48
'x' is the easiest to check here.. just add the digits of the number, sum is 26, thus this is not a multiple of 3.. since 'x' is an integer it must be '0'..
now neither of 'y' or 'z' could be '0' because just by looking we see that this number has trailing zeroes so it must have both 2 and 5...

so simply since 'x' is '0' the given solution is '0'. Hence B
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Re: For integers x, y, and z, if (3^x) (4^y) (5^z) = 3,276,800,0 [#permalink]

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Jut observe the RHS, 3,276,800,000 can be written as 32768 x 100000

32768 is not divisible by 3, so x = 0

xy/z = 0 = Answer = B; no need to calculate the rest
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Re: For integers x, y, and z, if (3^x) (4^y) (5^z) = 3,276,800,0 [#permalink]

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Re: For integers x, y, and z, if (3^x) (4^y) (5^z) = 3,276,800,0   [#permalink] 05 Sep 2015, 05:31
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For integers x, y, and z, if (3^x) (4^y) (5^z) = 3,276,800,0

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