BdMO National Secondary (Higher Secondary) 2011/6
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.
$p$ 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.
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Re: BdMO National Secondary (Higher Secondary) 2011/6
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor
- FahimFerdous
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Re: BdMO National Secondary (Higher Secondary) 2011/6
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
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
Your hot head might dominate your good heart!
Re: BdMO National Secondary (Higher Secondary) 2011/6
Fahim, Well done. Your proof is correct.
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor
- FahimFerdous
- Posts:176
- Joined:Thu Dec 09, 2010 12:50 am
- Location:Mymensingh, Bangladesh
Re: BdMO National Secondary (Higher Secondary) 2011/6
Thanks.
I've solved some other problems. I'll post my solutions after my SSC exam today.
I've solved some other problems. I'll post my solutions after my SSC exam today.
Your hot head might dominate your good heart!
Re: BdMO National Secondary (Higher Secondary) 2011/6
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 $p=3$
One one thing is neutral in the universe, that is $0$.
Re: BdMO National Secondary (Higher Secondary) 2011/6
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]
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Re: BdMO National Secondary (Higher Secondary) 2011/6
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.
"Your mind is like this water, my friend. When it is agitated,it becomes difficult to see. But if you allow it to settle , the answer becomes clear." Master Oogway.