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A pass code is created by combining one-digit prime numbers. What is t

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A pass code is created by combining one-digit prime numbers. What is t  [#permalink]

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New post 13 Jun 2019, 00:27
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Re: A pass code is created by combining one-digit prime numbers. What is t  [#permalink]

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New post 18 Jul 2019, 19:34
pushpitkc wrote:
There are 4 one digit prime numbers - 2,3,5,7.

The total number of 4 digit numbers possible is 4*4*4*4(since digits can be repeated)
In order to determine how many of these 4 digit numbers are divisible by 12,
we employ the following method :
For a number to be divisible by 12, it has to be divisible by both 3 and 4.

Only numbers ending with 32,52 and 72 can be divisible by 4
as the rule for divisibility of 4 is that the last 2 digits must be divisible by 4.
For the number to be divisible by 3, the sum of the digits must add to give a multiple of 3 :

We have the possible total combinations for the numbers ending :

1. _ _ 32 are 2532,5232,5532,3732,7332, and2232(6 combinations)
2. _ _ 52 are 7752,3552,5352,2352, and 3252(5 combinations)
3. _ _ 72 are 5772,7572,2772,7272, and 3372(5 combinations)

P(Numbers divisible by 12) = P(A) = 6+5+5 = 16
Also P(T) = P(Total possible combinations) = 256

The probability is \(\frac{P(A)}{P(T)}\) = \(\frac{16}{256} = \frac{1}{16}\) (Option A)


Hi there,
I notice that this is an old thread, but anyway I more or less solved the question the same way as you did, but I feel it's too time consuming and error-prone, especially when it gets to the part of divisible by 3. Basically I had to try all the combinations of numbers to see which sum works and it took me probably a good 10 minutes to make sure I didn't miss or over-count any combinations.

So is there a faster way to solve this within 2 min? How did you count the possible combinations of sums (divisible by 3)?
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Re: A pass code is created by combining one-digit prime numbers. What is t  [#permalink]

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New post 19 Jul 2019, 00:48
Bunuel
Can you help explain
why its only 16
--32 where first two can actually tally take 4*4 values =16 values
and thus for rest 2 also 16
thus total should be 16*3 = 48
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A pass code is created by combining one-digit prime numbers. What is t  [#permalink]

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New post Updated on: 19 Jul 2019, 20:25
As the number must be divisible 4, last 2 digits must be divisible by 4
1. ab32
2. ab52
3. ab72

32=1 mod 3, 52=2 mod 3, and 72=0mod3. As the last 2 digits, when divided by 3, in 3 cases gives different and all possible remainders, starting 2-digits, hence, cover all possible combinations of 2-digits that can be formed by 4 prime numbers.


Total 2 digits number than can be formed by the use of 2, 3, 5 and 7, when repetition is allowed= 4*4=16

Probability= 16/256=1/16



Bunuel wrote:
A pass code is created by combining one-digit prime numbers. What is the probability that the resulting four-digit number is a multiple of 12 ?

(A) 1/16
(B) 21/256
(C) 3/16
(D) 1/12
(E) 15/16

Originally posted by nick1816 on 19 Jul 2019, 18:48.
Last edited by nick1816 on 19 Jul 2019, 20:25, edited 1 time in total.
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Re: A pass code is created by combining one-digit prime numbers. What is t  [#permalink]

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New post 19 Jul 2019, 19:00
xx32... xx xan't take all 16 values, because sum of 4 digits must be divisible by 3.
32 leaves 2 remainder when divided by 3, hence xx must leave 1 remainder when divided by 3.

Hence in this case xx can be combination of 22, 25, 55 or 73.
Total arrangements possible= 1+2+1+2=6.

Similarly you have to find for the rest of the 2 cases.


globaldesi wrote:
Bunuel
Can you help explain
why its only 16
--32 where first two can actually tally take 4*4 values =16 values
and thus for rest 2 also 16
thus total should be 16*3 = 48
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Re: A pass code is created by combining one-digit prime numbers. What is t   [#permalink] 19 Jul 2019, 19:00
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