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Prime Time Revisited (Posted on 2004-10-22) Difficulty: 2 of 5
Do there exist three 2-digit primes such that:
  • Any two of the three, averaged, produce another prime, and
  • The average of all three is prime

See The Solution Submitted by SilverKnight    
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Solution Solution to "Prime Time Revisited" | Comment 15 of 17 |
 

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 possess 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)..(*).  Again. 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 wo 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 priome, 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.        


  Posted by K Sengupta on 2005-08-29 08:03:55
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