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by agps » Mon Aug 27, 2007 1:34 am
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.
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by beny » Mon Aug 27, 2007 3:00 am
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. :wink:
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Re: probability

by gabriel » Mon Aug 27, 2007 5:22 am
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 :) ..
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