All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Getting Primed With Subtraction And Addition (Posted on 2007-12-26)
Determine all possible primes p such that each of p-8, p-4, p+8 and p+12 are also primes.

 See The Solution Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 solution slightly differently | Comment 4 of 5 |
p, p-4 and p-8 are mutually unequal mod 3 (same as p, p-1, p+1) so at least one of these is a multiple of three. But if they're all primes, then the "multiple" of three must be 3 itself. p or p-4 can't be three or p-8 is negative, so p-8=3 is the only solution, resulting in p=11.

As a check, if p=11, p-4=7, p-8=3, and p+12=23 and indeed all are prime. Knowing that p+12 is prime is irrelevent to finding the solution, although it might have eliminated the one possibility had the condition failed to hold.

 Posted by Paul on 2007-12-27 21:05:57

 Search: Search body:
Forums (0)