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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

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

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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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