Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

This problem looks intimidating at first, but actually is [#permalink]
11 Nov 2005, 06:15

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

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!

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??

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.

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???

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!

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

Re: PS Statistics #22 [#permalink]
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!

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??

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. _________________

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 _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

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

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...