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A pass code is created by combining one-digit prime numbers. 23 Feb 2006, 07:56
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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?
a)1/16
b)21/256
c)3/16
d)1/12
e)15/16



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A pass code is created by combining one-digit prime numbers. 01 Apr 2006, 05:45
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The Answer is D...

the prime nos are 1, 2, 3, 5, 7

So the max combinations is 5! = 1x2x3x4x5 = 120

Prob shud be X/120..

By back solving... only D has the denominator as a factor of 120.

Whats the OA?
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A pass code is created by combining one-digit prime numbers. 01 Apr 2006, 07:33
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sm176811
The Answer is D...
the prime nos are 1, 2, 3, 5, 7
So the max combinations is 5! = 1x2x3x4x5 = 120
Prob shud be X/120..
By back solving... only D has the denominator as a factor of 120.
Whats the OA?
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1 is not a prime.
single digit primes are : 2, 3, 5 and 7. none of their combination is divided by 12.
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A pass code is created by combining one-digit prime numbers. 01 Apr 2006, 08:44
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I did a little calculation and it took around 30 seconds. I think the probability is 0.

First, the unit digit prime are 2,3,5,7 so that there are 24 combinations of password.

Among these, we need to identify combinations divisible by 12.

Since 12 is even number, the 4 digit passwor should end up with even number. In this case it's 2 only.

There are 3, 5, 7 left and we can make 6 4-digit number with 2 its unit digit.
(since I don't know general rule about numbers divisible by 12 I did a little calculation)
The 6 4-digit numbers are

3572
3752
5732
5372
7532
7352

nothing is divisible by 12.

In summary, the probability is 0
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A pass code is created by combining one-digit prime numbers. 01 Apr 2006, 08:50
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antiant
I did a little calculation and it took around 30 seconds. I think the probability is 0.

First, the unit digit prime are 2,3,5,7 so that there are 24 combinations of password.

Among these, we need to identify combinations divisible by 12.

Since 12 is even number, the 4 digit passwor should end up with even number. In this case it's 2 only.

There are 3, 5, 7 left and we can make 6 4-digit number with 2 its unit digit.
(since I don't know general rule about numbers divisible by 12 I did a little calculation)
The 6 4-digit numbers are

3572
3752
5732
5372
7532
7352

nothing is divisible by 12.

In summary, the probability is 0
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BTW, does anyone know a general rule regarding numbers divisible by 12?
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A pass code is created by combining one-digit prime numbers. 01 Apr 2006, 09:36
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The answer is A.

The question doesn't say that each prime digit can only appear once,
so there are |{2,3,5,7}|^4 = 4^4 = 256 possibilities beginning with
2222 and ending with 7777.

Much more difficult is determining the numbers that are divisible
by 12. A number is divisible by 12 if it's divisible by both 3 and 4.
So we can through the combinations where the checksum isn't
divisible by 3 and where the last 2 digits aren't divisible by 4
(which means that the whole number is divisible by 4).

Remaining are the following numbers satisfying the given condition:

2232
2352
2532
2772
3252
3372
3552
3732
5232
5352
5532
5772
7272
7332
7572
7752

Since these are 16 numbers, the answer to the question is
16/256 = 1/16 --> A.

But I guess very few could solve this in about 2 minutes (I didn't) :(
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A pass code is created by combining one-digit prime numbers. 01 Apr 2006, 17:26
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ccax
The answer is A.

The question doesn't say that each prime digit can only appear once,
so there are |{2,3,5,7}|^4 = 4^4 = 256 possibilities beginning with
2222 and ending with 7777.

Much more difficult is determining the numbers that are divisible
by 12. A number is divisible by 12 if it's divisible by both 3 and 4.
So we can through the combinations where the checksum isn't
divisible by 3 and where the last 2 digits aren't divisible by 4
(which means that the whole number is divisible by 4).

Remaining are the following numbers satisfying the given condition:

2232
2352
2532
2772
3252
3372
3552
3732
5232
5352
5532
5772
7272
7332
7572
7752

Since these are 16 numbers, the answer to the question is
16/256 = 1/16 --> A.

But I guess very few could solve this in about 2 minutes (I didn't) :(
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You are right, dude. There is not such line prime number can appear only once.

In the second part of your explanation, I would sugget you try my way.

First, all the numbers should finish with 2 b/c it's divisible by even number

Second, the last two digit should be divisible by 4.

put first&second condition together, there are OO32, OO52,OO72

finally, the checksum should be divisible by 3.

I don't know how long it will take but we can save some time by being a little more logical.

Thank you.
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A pass code is created by combining one-digit prime numbers. 01 Apr 2006, 18:12
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My Mistake... 1 is not prime! :p



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