BdMO National Secondary (Higher Secondary) 2011/6

[Moon]

Problem 6:
$p$ is a prime and sum of the numbers from $1$ to $p$ is divisible by all primes less or equal to $p$. Find the value of $p$ with proof.

[Avik Roy]

Once upon a time, there lived a man called Euclid in a beautiful country called Greece, who proved that there are infinitely many primes

[FahimFerdous]

I solved it. I can be wrong though. If I'm wrong, then plz forgive me and correct me. And I think there are easier solutions than mine. Here's my solution:

The summation of 1 to P is P(P+1)/2. Suppose it's equal to X. Now as X is divisible by every prime less than P, then they all must divide (P+1)/2 as P is prime. Now, as they're all co-prime, their product must divide (P+1)/2.
Suppose the product of all primes less than P is M. Now, we can easily prove that, (P+1)/2<P when P>1. It means that M must be less than P if it divides (P+1)/2.
Now, we suppose that M<P. Then, M+1 is also a prime because no prime less than P can divide it. Now, we can say that, either M>P or if M<P then M=P-1. Now, if (P+1)/2<P-1, then there's no solution. Because, then M can't divide it. So, (P+1)/2=P-1. Now, from this equation, we can show that, P=3. So, the only solution is 3.

Is my solution correct? :-S

[Avik Roy]

Fahim, Well done. Your proof is correct.

[FahimFerdous]

Thanks.
I've solved some other problems. I'll post my solutions after my SSC exam today.

[Masum]

Let $p_i$ be the primes,for $2\prod p_i|p_n+1=> p_n>2\prod p_i$ but for $i>3,p_n^2<p_1p_2.....p_{n-1},$so check $i=1,2,3$ which easily gives the solution $p=3$

[mahathir]

at first, we find the summation upto p. now, the summation upto p is p(p+1)/2. now, all primes before p must divide (p+1)/2. Now from (p+1)/2 to p,if there are any prime numbers, (p+1)/2 will not be divisible by them. Now except for p=5,3,2. for all other values of p, there will be primes between (p+1)/2 and p.Now, by putting the values p=2,3,5 we see that, p=3 is the solution and only solution.[the logic that there will be a prime between all other p and (p+1)/2 is that the gap between two primes q and p other than 2,3,5 is obviously smaller than the gap between p and (p+1)/2.because, at first we consider 7, we get then (p+1)/2 is 4. there is 5 between them.According to bertrand's postulate,there must be a prime number between p and p/2.since p/2 is not a natural number, then between p and (p+1)/2]

[Ibraheem Moosa]

The summation is p(p+1)/2. Dividing it by p we get (p+1)/2 which by assumption can be divided by every prime less than p.Now there is a prime number between (p-1) and (p-1)/2(As there is always a prime between a and 2a).This prime must divide (p+1)/2.So that prime can only be (p+1)/2.There is only one prime less than p.So p is the 2nd prime.So p is 3.