| View previous topic :: View next topic |
| Author |
Message |
gmatquestion
Guest
|
 Posted: Sun Aug 26, 2007 12:50 pm Post subject: probability |
 |
|
| A pass code is created by combining one-digit prime numbers. What is the probability that the resulting four-digit number will be a multiple of 12? |
|
| Back to top |
|
 |
|
|
agps
Really wants to Beat The GMAT!
Joined: 07 Aug 2007
Posts: 101
Thanks given: 0
Thanked 0 times in 0 posts
|
| Posted: Mon Aug 27, 2007 1:34 am Post subject: |
|
|
Hi.
there are 4 prime numbers with 1 digit 2,3,5,7.
so the total number of 4 digit combinatins with this numbers is 4! = 24;
you only want those that are divisible by 12 (or the sum of digits is divisible by 3 and last 2 digits multiple of 4)
for all combinations the sum of digist is 2+3+5+7 = 17, not a multiple of 3, therefore not a multiple of 12.
Probability 0.
this is not a result i was expecting, maybe i misread something in the question, lets wait for a 2nd opinion. |
|
| Back to top |
|
|
beny
Really wants to Beat The GMAT!
Joined: 20 Jul 2007
Posts: 214
Thanks given: 0
Thanked 2 times in 2 posts
|
| Posted: Mon Aug 27, 2007 3:00 am Post subject: |
|
|
| agps wrote: | Hi.
there are 4 prime numbers with 1 digit 2,3,5,7.
so the total number of 4 digit combinatins with this numbers is 4! = 24;
you only want those that are divisible by 12 (or the sum of digits is divisible by 3 and last 2 digits multiple of 4)
for all combinations the sum of digist is 2+3+5+7 = 17, not a multiple of 3, therefore not a multiple of 12.
Probability 0.
this is not a result i was expecting, maybe i misread something in the question, lets wait for a 2nd opinion. |
Hi agps,
You forgot that numbers can be re-chosen (7332 works). I do not have time to work out the problem right now, as I'm about to start class... but please re-think it.  |
|
| Back to top |
|
|
gabriel
Managing Director
Joined: 20 Dec 2006
Posts: 862
Thanks given: 118
Thanked 27 times in 26 posts
Location: India
|
| Posted: Mon Aug 27, 2007 5:22 am Post subject: Re: probability |
|
|
| gmatquestion wrote: | | A pass code is created by combining one-digit prime numbers. What is the probability that the resulting four-digit number will be a multiple of 12? |
Ok, the one digit prime numbers are 2,3,5,7 ... the number of ways you can create a 4 digit number without any restriction is 4*4*4*4 = 4^4 ..
.. Now we need to find the number of ways in which a number divisible by 12 can be made using these 4 numbers ..
..A few things to remember
.. for a number to be divisible by 12 it should be divisible by 3 and 4 ...
..For the number to be divisible by 4 the last two digits should be divisible by 4..
.. for a number to be divisible by 3 the sum of the digits should be divisible by 3...
using the divsibility test for 4 we know that out of the 4 numbers only 32,52 and 72 can occupy the last 2 places ..
for the number ending with a 32 ... the possible combinations are3732,7332,2232,2532,5232,5532... that is 6 in all ..
For the number ending with 72 .. the possible combinations are 2772,7272,3372,7572,5772... that is 5 in all ..
For the number ending with 52 .. the possible combinations are 2352,3252,3552,5352,7752, ... 5 in all
So the total = 16.. the probability = 16/4^4 = 1/16 .... hope i got all the numbers  .. |
|
| Back to top |
|
|
|
|
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum
You cannot attach files in this forum
You cannot download files in this forum
|
|
"GMAT" and other GMAC™ trademarks are registered trademarks of the Graduate Management Admission Council™. The Graduate Management Admission Council™ does not endorse, nor is it affiliated in any way with the owner or any content on this website. The opinions expressed here are solely those of the author or those of the members of this website. Copyright © 2008 BTG Test Prep, LLC. Powered by phpBB © 2001, 2005 phpBB Group.
|
|
