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

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

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.

The 4-digit passcode is created using 2,3,5,7.

(A) Number of possible passcodes = 4 x 4 x 4 x 4 = 256 (for each of the 4 digits, any of the 4 prime numbers are possible - repeats allowed).

Multiple of 12 => multiple of both 3 and 4.

(1) Multiple of 4 is only possible if the passcode is even. Thus, the number must be in the form _ _ _ 2. The last digit cannot be 3,5, or 7.

(2) Multiple of 4 is only possible if the last two digits of the passcode present a number divisible by 4. Thus,

-> _ _ 3 2
-> _ _ 5 2
-> _ _ 7 2

But not _ _ 2 2.

Now, we also have to ensure that the passcode is a multiple of 3. For this to happen, the sum of all the digits must be a multiple of 3.

Here, it makes sense to check - case by case.

(1) a b 3 2

A general observation: Given the numbers 2,3,5, and 7, the least possible a + b = 2+2 = 4, and the highest possible a + b = 7+7 = 14.

Now,

3+2 = 5. Now, how can we get 3-multiples from here?

-> a+b = 4. (2,2)
-> a+b = 7. (2,5) (5,2)
-> a+b = 10. (5,5) (3,7) (7,3)
-> a+b = 13. Not possible

Total: 6 possible numbers of form a b 3 2.

(2) a b 5 2

5+2 = 7. Similar to the above, how can we get 3-multiples from here?

-> a+b = 5. (2,3) (3,2)
-> a+b = 8. (5,3) (3,5)
-> a+b = 11. Not possible
-> a+b = 14. (7,7)

Total: 5 possible numbers of form a b 5 2.

(3) a b 7 2

7+2 = 9. Similar to the above, how can we get 3-multiples from here?

-> a+b = 6. (3,3)
-> a+b = 9. (2,7) (7,2)
-> a+b = 12. (5,7) (7,5)

Total: 5 possible numbers of form a b 7 2.

(B) Number of passcodes which are multiples of 12 = 6+5+5 = 16

Required Probability = B/A = \(\frac{16}{256}\) = 1/16

___
Harsha

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 ?

One digit prime numbers = {2,3,5,7}

Total ways to create a 4-digit code = 4^4 =256

For the code to be a multiple of 12, it should be divisible both by 4 & 3.

Numbers ending with 32, 52, 72 are divisible by 4.

For the code to be divisible by 3, the sum of its digits should be divisible by 3.

Possible codes = {2532, 5232, 2232, 5532, 7332, 3732, 2352, 3252, 5352, 3552, 7752, 3372, 7272, 2772, 7572, 5772}

Total number of favourable ways = 16

The required probability = 16/256= 1/16

IMO A

divisible by 12 means, it has to be divisible by 4 and also 3.

First, checking for divisible by 4 is easy and narrows the options, (to be divisible by 4, the last two digits have to be divisible by 4), so the options are:
_,_,32
_,_,52
_,_,72

Out of these, to be divisible by 3, each of them have some deficits:
_,_,32 -> sum is 5 -> so need 1
_,_,52 -> sum is 7 -> so need 2
_,_,72 -> sum is 9 -> so need 0

Now, for the first two places, no matter whichever combination you take, if you do %3 for their sum, they have to give back 0, 1 or 2 as remainder. (which is obvious / trivial from number theory that when you take a number and divide by n, the possible remainders are 0...n-1) for ex:
let's say you took, 57 for first two positions: 5+7 is 12 -> so remainder with 3 is 0
let's say you took, 23 for first two positions: 2+3 is 5 -> so remainder with 3 is 2
let's say you took, 55 for first two positions: 5+5 is 10 -> so remainder with 3 is 1

Now, the spark moment:
if you take any two single digit primes for first two spots -> they will give a remainder between (0, 1 and 2) when divided by 3
if you take the 3 numbers which passed the divisible by 4 criteria for the last two spots -> they have either a deficit of (0, 1 or 2)

thus, you can match all the first two spots with one of the last two spots that are divisible by 4

For ex: take a random fill for the first two digits:
say 23 -> sum is 5 -> remainder with 3 is 1. So it has 1 extra to donate. So it will fit with _,_,32 (since for _,_,32 -> sum is 5 -> so needs 1 to be divisible by 3)

another ex:
say 77 -> sum is 14 -> remainder with 3 is 2. So it has 2 extra to donate. So it will fit with _,_,52 (since _,_,52 needs 2 to be divisible by 3)

another ex:
say 33 -> sum is 6 -> remainder with 3 is 0. So it has nothing extra to donate. So it will fit with _,_,72 (since _,_,72 doesn't need anything to be divisible by 3)

So this way all the combinations for the first two places will fit in without overlapping nor without missing anything with one of the last two spots that are divisible by 4

Since repetition is allowed, number of combinations for first two places would therefore be:
4 * 4 => numerator (again all of these 16 numbers would fit with one of the _,_,32 or _,_,52 or _,_,72)

all possibilities with repetition allowed:
4 * 4 * 4 * 4 => denominator

ans: 1/16


It is a very hard problem. I only got this revelation after trying to understand the pattern in the solution and also the goal of how can we solve this in under 2 minutes without listing out all the numbers manually.

This is how I did it and it was a bit faster:
we know that the number must be divisible by 3 and 4 to be divisible by 12 (because 3*4 = 12)
we also know the last digit has to be 2 since it must be an even number
x,x,x,2
we can write out the possible last 2 digit combinations
x,x,22
x,x,32
x,x,52
x,x,72
and 72 is the only number divisible by 3 and 4.

So we have 4*4*1*1 many possibilities out of 4^4
which gives you 1/16

sriharsha4444

Thanks, I think this is the necessary insight in order to be able to solve in 2 minutes without luck. Great work finding the logic there. Amazing that you have to stop thinking about the last 2 numbers in order to find the realisation that each pair of ab corresponds to exactly one of the options for cd, so you can include all permutations for ab. Nice.