comparing primes

[phil1882]

inspired by bonanova's puzzle, here's mine.

there are three boxes each containing a prime. it is known that the largest of the three is no more than twice the size of the smallest. (so 7, 11, 13 is the smallest possible values for the three boxes.)

you open one box, and must guess whether your number is smaller, larger or in between the other two values.

which should you guess?

 

 

[karthickgururaj]

inspired by bonanova's puzzle, here's mine.

there are three boxes each containing a prime. it is known that the largest of the three is no more than twice the size of the smallest. (so 7, 11, 13 is the smallest possible values for the three boxes.)

you open one box, and must guess whether your number is smaller, larger or in between the other two values.

which should you guess?

One can enumerate the possible options after seeing the number in the first box?

 

For example, say the number in the first box is 31.

 

If 31 is to be the smallest of the lot, then the largest can be max 61 and min 41, so we have the following possibilities:

If largest is 41: (31, 37, 41) -> 1 possibility

...

If largest is 61: (31, 37 or 41 or 43 or 47 or 53 or 59 , 61) -> 6 possibilities.

 

The total number of possibilities is: 1 + 2 + 3 + 4 + 5 + 6 = 21.

 

Or in general,

1. If p₁ is the number in the first box

2. Let p₂, p₃, etc be the next prime numbers that occur in the integer sequence after p₁.

2. Let p_n be the highest prime that is less than 2*p₁

3. The number of possibilities where p₁ happens to be the smallest is: (n-2)*(n-1)/2

 

Next, if the number is to be the middle prime, again taking 31 as an example, the possible values are:

(19 or 23 or 29, 31, 37) -> 3 possibilities

(23 or 29, 31, 41) -> 2 possibilities

(23 or 29, 31, 43) -> 2 possibilities

(29, 31, 47) -> 1 possibility

(29, 31, 53) -> 1 possibility

 

Total: 9 possibilities.

 

Slightly harder to express this in general terms..

 

Finally, if the number is to be the largest, then we have the following possible triplets:

(17 or 19 or 23, 29, 31) -> 3 possibilities

(17 or 19, 23, 31) -> 2 possibilities

(17, 19, 31) -> 1 possibilities

 

Total: 1 + 2 + 3  = 6 possibilities.

 

Just like the first case (where p₁ was to be the smallest), it can be expressed in slightly more general terms..

1. If p₁ is the number in the first box

2. Let p₂, p₃, etc be the previous prime numbers that occur in the integer sequence before p₁.

2. Let p_n be the lowest prime that is greater than p₁/2

3. The number of possibilities where p₁ happens to be the largest is: (n-2)*(n-1)/2

 

We can go with the choice with the largest number of possibilities..