Well, this problem continues to interest me. Picking up on a Charlie's idea, let a = pk + b where k and b are integers and 0 <= b < p. We potentially need to consider all 12 cases of k mod 12 (because 12 is the LCM of 1,2,3 and 4), but we know that a < 9p (see previous post), so we only need to consider k = 0 through 8, and not worry about mod 12 at all. Just 9 cases is an improvement over the prior methodology, and here they are:

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**k = 0 (ie, a < p)**

b + b + b + b = (b) + p

b = p/3

works for p = 3, b = 1, a = 1

**Solution a = 1, p = 3**

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k = 1

b + (p+b) + (p+b) + (p+b) = (p + b) + p

b = -p/3

No solution, since b is negative

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k = 2

b + (b) + (2p+b) + (2p+b) = (2p + b) + p

b = -p/3

No solution, since b is negative

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**k = 3 **

b + (p+b) + (b) + (3p+b) = (3p + b) + p

b = 0

**Solution a = 3p **

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k = 4

b + (b) + (p+b) + (b) = (4p + b) + p

b = p*(4/3)

No solution, since b > p

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**k = 5 **

b + (p+b) + (2p+b) + (p+b) = (5p + b) + p

b = p*(2/3)

works for p = 3, b = 2, a = 17

**Solution a = 17, p = 3**

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k = 6

b + (b) + (b) + (2p+b) = (6p + b) + p

b = p*(5/3)

No solution, since b > p

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k = 7

b + (p+b) + (p+b) + (3p+b) = (7p + b) + p

b = p

No solution, since b = p

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k = 8

b + (b) + (2p+b) + (b) = (8p + b) + p

b = p*(7/3)

No solution, since b > p

*Edited on ***December 31, 2012, 11:28 am**