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All that remainder is Gold (Posted on 2019-07-21) Difficulty: 3 of 5
(n3-12n2+8n-93)/(11+2n-n2) = p

Let n be some integer satisfying the above equation yielding a prime number p. What is the sum of all possible values of p.

No Solution Yet Submitted by Danish Ahmed Khan    
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No Subject | Comment 1 of 3
(n^3-12n^2+8n-93)/(11+2n-n^2)=
10-n+(n+203)/(n^2-2n-11)
so for this to be an integer we need
n+203>=n^2-2n-11
which for integer n is only true when
-13<=n<=16
from this we get that the only integer solutions are
(n,p):
(-3,63)
(-2,-55)
(1,-8)
(4,-63)
(5,57)

From these, the only one where p is prime is when n=5 which gives p=57.

Thus the only possible prime value for p is 57 and thus the sum is 57.

  Posted by Daniel on 2019-07-22 09:14:38
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