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Getting Primed With Subtraction And Addition (
Posted on 20071226
)
Determine all possible
primes
p such that each of p8, p4, p+8 and p+12 are also primes.
Submitted by
K Sengupta
No Rating
Solution:
(
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Only p=11 satisfies the conditions of the problem.
EXPLANATION:
Let us substitute p8 = q, so that:
(p8, p4, p+8, p+12) = (q, q+4, q+16, q+20)
If q (Mod 3) = 0, then q is non prime unless q=3
If q (Mod 3) = 1, then q+20 is divisible by 3, an hence composite. This is a contradiction.
If q(Mod 3) = 2, then each of q+4 and q+16 is divisible by 3, and hence both are composites. This is a contradiction.
Accordingly, q=3 and consequently, p8=3, so that p=11 is the only possible solution to the given problem.
*** Also refer to the solution submitted by
Paul
in this
location
.
Comments: (
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)
Subject
Author
Date
re: solution slightly differently
K Sengupta
20080306 04:32:20
solution slightly differently
Paul
20071227 21:05:57
Solution
Praneeth
20071227 00:49:41
solution/spoiler
xdog
20071226 12:28:44
computer exploration (spoiler)
Charlie
20071226 11:59:39
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