BdMO National 2013: Secondary 9, Higher Secondary 7

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BdMO
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BdMO National 2013: Secondary 9, Higher Secondary 7

Unread post by BdMO » Fri Jan 10, 2014 1:40 am

If there exists a prime number $p$ such that $p+2q$ is prime for all positive integer $q$ smaller than $p$, then $p$ is called an "awesome prime". Find the largest "awesome prime" and prove that it is indeed the largest such prime.

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Fatin Farhan
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Re: BdMO National 2013: Secondary 9, Higher Secondary 7

Unread post by Fatin Farhan » Sun Jan 12, 2014 5:50 pm

If $$p=2$$ then $$p+2q \equiv 0 (mod 2)$$
So $$p \neq 2$$.
If $$p=3$$ then $$2q=2,4$$
So, $$p+2q=5,7$$
So, $$3$$ is a awsome prime.
If $$p>3$$ then there are q's such that $$2q \equiv 0,1,2 (mod 3)$$.
Now, if $$p \equiv 1,2(mod 3)$$
So $$p+2q \equiv 0 (mod 3) $$ for some q's.
So $$p \ngtr 3$$.
So $$3$$ is the largest $$awsome-prime$$.
"The box said 'Requires Windows XP or better'. So I installed L$$i$$nux...:p"

shahriarabdullah
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Re: BdMO National 2013: Secondary 9, Higher Secondary 7

Unread post by shahriarabdullah » Fri Feb 07, 2014 1:48 am

i p=2 then $2|p+2q$. So p cannot be 2.
Let p be the largest awesome prime.
The least value of q for any p is 1. So, we can write-

$ p+2\equiv1,2 (mod 3) $
now suppose,
$ p+2\equiv1 (mod 3) $
so, $p+4\equiv0 (mod 3) $
so, $2q=4$
and so, $q=2$. which is not possible.

So, for the largest awesome prime-
$ p+2\equiv2 (mod 3) $
or, $ p+6\equiv0 (mod 3) $
or, $ p\equiv0 (mod 3)$
Now we can say, p is a prime and divisible by 3. So the only value of $p=3$

And it's the largest awesome prime. :)

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