Advanced Math/Primes

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I can't figure this one out. These types of problems always give me trouble.

Four positive integers are represented by x, x+d, x+2d, and x+3d where 4x+6d<150. If d is a positive integer and x+1, x+d+1, x+2d+1, and x+3d+1 are all positive prime integers, find the sum of all distinct possibilities for x.

Also, thank you for your insight on the previous problem I sent you. I found it very helpful. Thanks again.

Sincerely,
Isaac

Answer
Questioner:   Isaac Hawn
Country:   Illinois, United States
Category:   Advanced Math
Private:   Yes <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<< changed
Subject:   difficult algebra problem
Question:   I can't figure this one out. These types of problems always give me trouble.

Four positive integers are represented by x, x+d, x+2d, and x+3d where 4x+6d<150. If d is a positive integer and x+1, x+d+1, x+2d+1, and x+3d+1 are all positive prime integers, find the sum of all distinct possibilities for x.

Also, thank you for your insight on the previous problem I sent you. I found it very helpful. Thanks again.

Sincerely,
Isaac
...............................................
x, x+d, x+2d, and x+3d

If x+1, x+1+d, x+1+2d, and x+1+3d are all positive prime integers then at most one of them can be even, i.e. equal to 2, so assume all are odd primes.

If 4x + 6d < 150, then the primes add to 4x+4+6d < 154.

Their average is <= 38.5
The sum of the middle two is <= 77
But they are both odd, so <= 76

Now I don't have any brilliant insight for you, except to work your way down:

Can two primes p2,p3 add to 76?
If so, p1 = p2 - (p3 - p2) = 2p2 - p3
and    p4 = p3 + (p3 - p2) = 2p3 - p2

p2 = 23,  p3 = 53  NO
p2 = 29,  p3 = 47  NO p1 = 11, p4 = 69 not prime
p2 = 23,  p3 = 53  NO

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Can two primes p2,p3 add to 64?
p2 = 23,   p3 = 41  Hmmm..   p1 = 23-18=5, p4 = 41+18=59

So... 5,23,41,59  is your arithmetic sequence.  x = 4, d = 18.

Finish up.

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