Search found 2 matches
- Fri Feb 07, 2014 1:48 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Secondary 9, Higher Secondary 7
- Replies: 3
- Views: 4393
Re: BdMO National 2013: Secondary 9, Higher Secondary 7
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-...
- Mon Jan 13, 2014 6:10 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2013: Junior 7
- Replies: 1
- Views: 3094
Re: BdMO National 2013: Junior 7
Let $x$ be the smallest Fantabulous number. Due to condition we know that it is a positive integer. The smallest positive integer is $1$. So $x=1$. All other Fantabulous numbers will be valid if $1$ is a valid Fantabulous number. But there is no smaller positive integers than $1$. So $1=x$ can not b...