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The purpose of this problem is to exploit a weakness used by [#permalink]
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17 Nov 2005, 23:11
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The purpose of this problem is to exploit a weakness used by ETS: complimentary answer choices. Almost always in complimentary probability questions, there are a pair of "LUCKY TWINS" among the answer choices. If in doubt and pressed for time, choose a TWIN by logical deduction.
Let`s take a crack at this Project GMAT bad boy without making lengthy calculations. Duttist, GSR, Titleist, Laxie, Gamjatang, Wilfred, Nakib, etc... NO PENCILS allowed! Please break out the big guns and explain your deductive thought process.
#50: Set S consists of numbers 2, 3, 6, 48, and 164. Number K is computed by multiplying one random number from set S by one of the first 10 nonnegative integers, also selected at random. If Z=6^K, what is the probability that 678,463 is not a multiple of Z?
a. 10%
b. 25%
c. 50%
d. 90%
e. 100%
*LUCKY TWINS



Director
Joined: 21 Aug 2005
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Well, my probability skills is not one that i can boast of
Looking at this, the number 678,463 seems to be a prime number (atleast not divisible by (2, 3, 4, 6, 7, 8, 9, 11...).
Also, 6^k is even.
On the exam I will go for E) 100%



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Could you elaborate a little more on your lucky twins and how you came to choose 10% and 90% as your twins?



Current Student
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ywilfred wrote: Could you elaborate a little more on your lucky twins and how you came to choose 10% and 90% as your twins?
.9+.1=1 (I like your humor Wilfred)



Current Student
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gsr wrote: Well, my probability skills is not one that i can boast of Looking at this, the number 678,463 seems to be a prime number (atleast not divisible by (2, 3, 4, 6, 7, 8, 9, 11...). Also, 6^k is even. On the exam I will go for E) 100%
E is not a LUCKY TWIN (no corresponding 0% among the answer choices) so no "Yukay" for you GSR.



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GMATT73 wrote: E is not a LUCKY TWIN (no corresponding 0% among the answer choices) so no "Yukay" for you GSR.
See I've become dumb after taking the exam!



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Joined: 22 Aug 2005
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90%
678,463 is weird number..very few numbers will be its multiple.
it cannot be 100% as K can be 0 and thus 6^0 = 1.
678,463 is multiple of 1
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Director
Joined: 14 Sep 2005
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duttsit wrote: 90%
678,463 is weird number..very few numbers will be its multiple.
it cannot be 100% as K can be 0 and thus 6^0 = 1. 678,463 is multiple of 1
At first, I thought it should be (E) 100% because 678463 is an odd number and Z is an even number.
However, there is a hidden 1 as duttsit explained.
K can be 50 different numbers, and in 5 cases K is 0.
Therefore, 1  5/50 = 90%
(D) should be it.
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Last edited by gamjatang on 18 Nov 2005, 00:38, edited 1 time in total.



SVP
Joined: 24 Sep 2005
Posts: 1885

well, let's see: because the second group contains the first 10 nonnegative integers ...there's the case of 0 to be picked ..and the probability of picking 0 from this group is 1/10
The number 6.... (hik,don't remember) can only be a multiple of Z in case Z=1 , that is to say K=0 ..since K is the product of the two stated groups...the prob of K be 0 is 1/10 or 10%
So to let the conditon satisfied , the prob blah blah is 100%10%=90%



Director
Joined: 14 Sep 2005
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Location: South Korea

GMATT73 wrote: gsr wrote: Well, my probability skills is not one that i can boast of Looking at this, the number 678,463 seems to be a prime number (atleast not divisible by (2, 3, 4, 6, 7, 8, 9, 11...). Also, 6^k is even. On the exam I will go for E) 100% E is not a LUCKY TWIN (no corresponding 0% among the answer choices) so no "Yukay" for you GSR.
Yukay is really expensive!!!
However, what is LUCKY TWINS?
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Auge um Auge, Zahn um Zahn !



SVP
Joined: 24 Sep 2005
Posts: 1885

gamjatang wrote: However, what is LUCKY TWINS?
for example: two answer choices X% and Y% , if X% + Y% =100% , they're lucky twins and possibly correct answer so pay much attention to these pairs. Those answer choices which can't find another choice to form lucky twins can be eliminated. Thanks to Matt for nice trick



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Re: The purpose of this problem is to exploit a weakness used by [#permalink]
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27 Feb 2014, 14:21
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Re: The purpose of this problem is to exploit a weakness used by
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