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This problem looks intimidating at first, but actually is

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This problem looks intimidating at first, but actually is [#permalink] New post 11 Nov 2005, 06:15
00:00
A
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C
D
E

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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
This problem looks intimidating at first, but actually is very easy once you apply the right strategy.. Set the two minute timer and attack! :war

A password to a certain database consists of digits that cannot be repeated. If the password is known to consist of at least 8 digits and it takes 12 seconds to try one combination, what is the amount of time, in minutes, necessary to guarantee access to the database?

A. 8!/5
B. 8!/2
C. 8!/12
D. 10!/2
E. 5/2*10!

Project GMAT divides this problem into 6 distinct steps. Any consolidation tips??
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 [#permalink] New post 11 Nov 2005, 06:22
What does at least 8 digits mean? is there a max of digites? I mean can there be a 20 digit code?

heeh, I got A, assuming the code simply has 8 ditinct digits. but in like one step and thats too easy to be a project GMAT problem :P
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 [#permalink] New post 11 Nov 2005, 06:31
Nero> hint: this is one of those "at least/at most" probability brain teasers. Chad and Markus rate this problem as being moderately difficult, but actually it`s much easier than you think. :boxer2
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 [#permalink] New post 11 Nov 2005, 06:57
As far as i understand this Q it should be 8!/12+9!/12+10!/12 , cause the digits can not be repeated or (100x8!)/720 or 5/36x8!
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 [#permalink] New post 11 Nov 2005, 07:01
I am not sure, let me try

I assume by digit you mean nuber digits. so we have 10 digits available including 0 1 2 3 4 5 6 7 8 9

if the password is consisting of at least 8 digits and digits can not be repeated,

we have 8 digit or 9 digit or 10 password
so we get 10 C 8 + 10C 9 + 10 C 10

= 45+ 10 +1
= 56 that is the total possible number of passwords.

we got 12 sec. to try one combination
so in minutes it will take 56/5 minutes to guarantee access.

I don't see any answer choice close to 56/5....

Did I make any mistake, Matt?
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 [#permalink] New post 11 Nov 2005, 07:18
So far Nakib is the closest. Hint #2: This is not a combination problem, rather a Permutation problem where the order of the digits is irrelevant.
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 [#permalink] New post 11 Nov 2005, 07:32
Is my assumption regarding digits incorrect?
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 [#permalink] New post 11 Nov 2005, 07:39
I can just feel it.. Somebody out there is going to have an sudden epiphenal "ah-ha" and crack the code any minute now... Nakib, Laxie, Gsr, Duttsit, Nero, Bumblebeeman, Titleist.. Who gets to savor the kanpai first??? :circle
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 [#permalink] New post 11 Nov 2005, 07:42
is it E ? ...there're 10 cases for 1st digit , 9 for 2nd digit , 8 for 3rd one, and so forth .....3 for last digit ---> 10*9*8*....*3= 10!/2
One combination in 12 second or 1/5 minute thus for 10!/2 combinations, we need 10!/2 / 1/5= 5/2*10!
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 [#permalink] New post 11 Nov 2005, 07:46
How bout this:
We need not try any 7 digit combinations. just all others:
8 digits, 9 digits and 10 digits (which is simply 10!)

The we got 10P8 + 10P9 + 10!
10!/2 + 10! + 10!
10!/2 + 2*10!
(10! + 2*2*10!)/2
5*10!/2
multiply by 12 and divide by 60 = divide by 5
10!/2

D?
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 [#permalink] New post 11 Nov 2005, 08:00
BULL`S EYE! Way to go Nero!! :cool

OA is D.

OE:

Since there are 10 distinct digits (0-9), none of which can be repeated, the password can contain at most 10 digits. Given that there are at least 8 digits in the password, it can contain 8, 9, or 10 digits. Because rearranging the digits leads to a new password, we can apply the permutations formula.

Number of 8 digit passwords: 10!/(10-8)!---->10!/2

Number of 9 digit passwords: 10!/(10-9)!---->10!

Number of 10 digit passwords: 10!/(10-10)!---->10!

Total number of possible passwords: 10!/2+10!+10!---->10!(.5+1+1)---->5/2*10!

Since it takes 12 seconds (1/5 of a minute) to try one password, we can calculate the time necessary to try all combinations: 5/2*10!*1/5---->10!/2
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Re: PS Statistics #22 [#permalink] New post 11 Nov 2005, 08:04
D.
at least 8 digits ...

total no of password permutation ( not combination )

10P8+10P9+10P10= 10!(1/2!+1+1)=10!*5/2

total time taken to try all the passwords = 10!*5/2*12/60=10!/2




GMATT73 wrote:
This problem looks intimidating at first, but actually is very easy once you apply the right strategy.. Set the two minute timer and attack! :war

A password to a certain database consists of digits that cannot be repeated. If the password is known to consist of at least 8 digits and it takes 12 seconds to try one combination, what is the amount of time, in minutes, necessary to guarantee access to the database?

A. 8!/5
B. 8!/2
C. 8!/12
D. 10!/2
E. 5/2*10!

Project GMAT divides this problem into 6 distinct steps. Any consolidation tips??
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 [#permalink] New post 11 Nov 2005, 08:18
:arh
I am tearing my hair.... I should shave my head....
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 [#permalink] New post 11 Nov 2005, 08:55
nakib77 wrote:
:arh
I am tearing my hair.... I should shave my head....

Just saw this problem. Pretty good done nabik77. You shown right direction :)
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 [#permalink] New post 25 Nov 2005, 20:55
nero44 wrote:
How bout this:
We need not try any 7 digit combinations. just all others:
8 digits, 9 digits and 10 digits (which is simply 10!)

The we got 10P8 + 10P9 + 10!
10!/2 + 10! + 10!
10!/2 + 2*10!
(10! + 2*2*10!)/2
5*10!/2
multiply by 12 and divide by 60 = divide by 5
10!/2

D?


Can you, plz, explain why we are adding 10P8 + 10P9 + 10!, because I thought it was supposed to be 10P8 * 10P9 * 10! which is the mess. I guess that, because first 8 digits stay the same. What's was I thinking??? thank you.
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 [#permalink] New post 26 Nov 2005, 11:11
It could be 8 digits (8P10), 9 digits(9P10) or 10 digits(10P10). So total outcomes are 8P10+9P10+10P10=10!/2+10!/1+10!=5/2*10!
Now each outcome needs 12/60=1/5 minutes.
So total minutes is 5/2*10!*1/5=10!/2
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 [#permalink] New post 26 Nov 2005, 11:30
this one is tough...

or maybe my mind is too slow...another week break...

OK lets see...

we have 10P8 for getting all the 8 digits, we can have 10P9 for 9 digits, and we can have 10P10 for all the 10 digits..

what i mean is clearly we dont need to look at it anything less than 8 cause we need at least 8 digits to get the password correct..so we cannot have 7 digits, or 6 etc... and digits dont repeat.. so we have the possibility that of the 10 digits we get the first 8 correct, then the first 9 correct and then all the 10 digits...

so all possible possibilities of getting the right passcode are:10P8 +10P9 +10P10.. 10!/2 +10! +10!=10!(.5+1+1)=10! (5/2) since it take 12/60 minutes to break each digit it takes 1/5 min per try...so

we have 10!(5/2)*1/5=10!/2
  [#permalink] 26 Nov 2005, 11:30
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