There are 21 two-digit primes. Choosing a set of 3 would yield a combination of 21!/(3!*18!) = 1330 possibilities. I was able to first narrow this down to 50 before doing some checks.
Primes are either of the form 3n+1 or 3n+2. I grouped the 21 primes into those categories (10 3n+1 and 11 3n+2). The average of 3 primes must be an integer. The only way this is possible is if we choose all 3 of them from the group of 3n+1 or 3n+2. There's no way for the average to be an integer if we choose 1 from one group and 2 from the other group.
Prime numbers are also in the form 4x+1 and 4x+3. If any two of them average to a prime, they must be in the same group or the average will be divisible by 2 (composite). So subdividing the two categories into these gives 4 groups.
3n+1 and 4x+1 = 12y+1
13, 37, 61, 73, 97
3n+1 and 4x+3 = 12y+7
19, 31, 43, 67, 79
3n+2 and 4x+1 = 12y+5
17, 29, 41, 53, 89
3n+2 and 4x+3 = 12y+11
11, 23, 47, 59, 71, 83
The three numbers must all come from the same group. So far, that means the average of any 3 will be an integer. And the average of any two will be odd. This narrows it down from 1330 to 50 possibilities.
The next step is to check to see which pairs yield prime averages. If it were to belong to the solution, there must be 3 numbers that show up at least twice. That's because the number must be used twice to check for an average. Consider x, y, z. (x+y)/2, (x+z)/2, and (y+z)/2 must be primes and x, y, and z show up twice each and must pair with each other.
Group 1 (12y+1 : 13, 37, 61, 73, 97)
Prime Averages: (13, 61), (13, 73), (37, 97), (61, 73), (61,97)
Eliminations
-37 shows up once only
-97 shows up once only after 37 is eliminated
Possibilities: (13, 61, 73)
Group 2 (12y+7 : 19, 31, 43, 67, 79)
Prime Averages: (19, 43), (19, 67), (31, 43), (43, 79), (67, 79)
Eliminations
-31 shows up once only
-19 not paired with 79 at all
-19 paired with 43, 67. (43, 67) not a pair
-79 paired with 43, 67. (43, 67) not a pair
Possibilities: None
Group 3 (12y+5 : 17, 29, 41, 53, 89)
Prime Averages: (17, 29), (17, 41), (17, 89), (29, 53), (29, 89), (41, 53), (53, 89)
Eliminations
-41 shows up exactly twice with 17 and 53, (17,53) is not a
pair so 41 is elimination
-17 shows up exactly twice with 29 and 89, (29, 89) is a pair
Possibilities: (17, 29, 89), (29, 53, 89)
Group 4 (12y+11 : 11, 23, 47, 59, 71, 83)
Prime Averages: (11, 23), (11, 47), (11, 71), (11, 83), (23, 59), (23, 71), (23, 83), (47, 59), (47, 71), (59, 83)
Eliminations
-47 shows up three times with 11, 59, 71. (59, 71) not a pair
but (11, 71) is a pair
-59 shows up twice exactly with 23, 83. (23, 83) is a pair
-71 and 83 both show up twice and pairs are formed
Possibilities: (11, 47, 71), (23, 59, 83), (11, 23, 71), (11, 23, 83)
From the 50 possibilities, we've narrowed it down to 7. Checking the average of the 3 is the last step:
Average (13, 61, 73) = 49
Average (17, 29, 89) = 45
Average (29, 53, 89) = 57
Average (11, 47, 71) = 43
Average (23, 59, 83) = 55
Average (11, 23, 71) = 35
Average (11, 23, 83) = 39
Of these 7 possibilities, only (11, 47, 71) yield a prime average.
Edited on October 22, 2004, 4:56 pm
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Posted by np_rt
on 2004-10-22 16:53:07 |
Brute Force Solution
