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Re: The purpose of this problem is to exploit a weakness used by [#permalink]
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) 8-)
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Re: The purpose of this problem is to exploit a weakness used by [#permalink]
gsr wrote:
Well, my probability skills is not one that i can boast of :roll:
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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Re: The purpose of this problem is to exploit a weakness used by [#permalink]
GMATT73 wrote:
E is not a LUCKY TWIN (no corresponding 0% among the answer choices) so no "Yukay" for you GSR.


See :roll: I've become dumb after taking the exam! :oops: :cry:
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Re: The purpose of this problem is to exploit a weakness used by [#permalink]
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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Re: The purpose of this problem is to exploit a weakness used by [#permalink]
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.

Originally posted by gamjatang on 18 Nov 2005, 00:33.
Last edited by gamjatang on 18 Nov 2005, 00:38, edited 1 time in total.
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Re: The purpose of this problem is to exploit a weakness used by [#permalink]
well, let's see: because the second group contains the first 10 non-negative 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%
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Re: The purpose of this problem is to exploit a weakness used by [#permalink]
GMATT73 wrote:
gsr wrote:
Well, my probability skills is not one that i can boast of :roll:
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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Re: The purpose of this problem is to exploit a weakness used by [#permalink]
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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