There are 3 sets of 4 primes of at most 2 digits with an interval of 6. These are (5, 11, 17, 23), (41, 47, 53, 59) and (61, 67, 73, 79).
As S = W + X + Y + Z = 2(A + B + C + D), we have the following values of S for each set:
( 5, 11, 17, 23) : 112
(41, 47, 53, 59) : 400
(61, 67, 73, 79) : 560
Of these values, only 400 is a square number, therefore:
A = 41
B = 47
C = 53
D = 59
W = 88 [41 + 47]
X = 100 [47 + 53]
Y = 112 [53 + 59]
Z = 100 [41 + 59]
S = 400 [88 + 100 + 112 + 100]
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Posted by Dej Mar
on 2009-08-24 11:41:23 |