September 2025 Problem of the Month

Determine three 2-digit primes such that:
Any two of the three, averaged, produce another prime, and
The average of all three is prime.

Solution to the Problem:

Here is K. Sengupta's solution:

THE REQUIRED THREE TWO DIGIT PRIMES ARE 11, 47 AND 71.

EXPLANATION : Let the three prime numbers be A,B and C.
Accordingly, each of A,B and C must possess either of the forms (6p+1) or (6p+5).
If one of the numbers possess the (6p+1) while one of the remaining numbers possesses the form (6p+5), then their average would possess the form (6p+3) which is divisible by 3 and consequently not a prime number.

Accordingly, all of the three numbers must possess the form (6p+1) or all of the three numbers must possess the form (6p+5)..(*****).
Since A,B and C are prime numbers they must possess a remainder of 1 or 3 or 5 or 7 upon division by 10.

The average of each of the pair of numbers is a prime, and accordingly, it follows that no two of the pairs of numbers can together possess the respective forms (10q+3 and 10q+7) or (10q+1 and 10q+9) since otherwise the average of the said pairs would be divisible by 5.

Since the average of the three numbers must be a prime, it follows that A,B and C must possess the respective forms ( 10q+1,10q+3 and 10q+3) or ( 10q+1,10q+1 and 10q+7) or ( 10q+3,10q+9 and 10q+9 ) or ( 10q+7,10q+7 and 10q+9) , since otherwise , the average of A,B and C would be divisible by 5.

Now any pair of numbers possessing the same remainder upon division by 10 must necessarily be separated by an even multiple of 10 or a multiple of 20 because if these numbers were separated by an odd multiple of 10 , then their average would be even, which contradicts the conditioin of the problem.

But, from (*****) , we know that such pairs must also differ by a multiple of 6.. consequently, any pair of numbers possessing the same remainder upon division by 10 must differ by LCM (20,6) = 60 or its multiple.

We observe that, there are four possible triplets of two digit primes given by (A,B,C) = (11,23,83), (13,61,73), (11,47,71), (23,53,89) such that the average of all the pairs in each triplet are primes.

Now, of the four triplets, only the triplet (A,B,C) = (11,47,71) satisfies the condition that the average of all the three numbers is a prime number.

Hence the required three 2-digit primes are 11, 47 and 71.
Check: the average of 11 and 47 is 29, a prime.
Check: the average of 11 and 71 is 41, a prime.
Check: the average of 71 and 47 is 59, a prime.
Check: the average of 11, 47 and 71 is 43, a prime.

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Sean Ulmer gets extra credit for finding 11, 47, and 71 and then also finding, "If the Solutions do not have to be unique I came up with 11, 23, 23."