Consider any set of 20 consecutive positive integer numbers greater than 50; what is the greatest possible
prime numbers in the set?
Mymensingh 2008 /5
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nafistiham
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Re: Mymensingh 2008 /5
could it be a little elaborated?
i am confused that does the problem want the greatest prime or the greatest possibility of having a prime in $20$ consecutive numbers greater than $50$.
i am confused that does the problem want the greatest prime or the greatest possibility of having a prime in $20$ consecutive numbers greater than $50$.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Mymensingh 2008 /5
By trial an error, I get 5 as an answer. (If I understand the problem right).
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Re: Mymensingh 2008 /5
You got that right Tiham...
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nafistiham
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Re: Mymensingh 2008 /5
as we know the greater primes get the difference between them increases.so, we just have to find the richest $20$ consecutive which are near $50$
my answer is $5$, too
my answer is $5$, too
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Mymensingh 2008 /5
I used the same idea...
Cut I think there's a prime number theorem to predict prime density....
can't use it though...
Cut I think there's a prime number theorem to predict prime density....
can't use it though...
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Abdul Muntakim Rafi
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Re: Mymensingh 2008 /5
We can't say this...
e.g.
$7507 , 7517 , 7523 , 7529 , 7537 , 7541 , 7547 , 7549 , 7559 ,7561,
7573 , 7577 , 7583 ,7589, 7591$ 15 prime numbers from 7500 to 7591
$7001 , 7013 ,7019 ,7027 , 7039 , 7043 , 7057 , 7069 ,7079$ 9 prime numbers From 6998 to 7103
I have made a progress... I will discuss this tonight... Have to go now...
e.g.
$7507 , 7517 , 7523 , 7529 , 7537 , 7541 , 7547 , 7549 , 7559 ,7561,
7573 , 7577 , 7583 ,7589, 7591$ 15 prime numbers from 7500 to 7591
$7001 , 7013 ,7019 ,7027 , 7039 , 7043 , 7057 , 7069 ,7079$ 9 prime numbers From 6998 to 7103
I have made a progress... I will discuss this tonight... Have to go now...
Man himself is the master of his fate...
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nafistiham
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Re: Mymensingh 2008 /5
wow.this is really amazing. 
but in the case of the question i think it can't cross $5$
but in the case of the question i think it can't cross $5$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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bristy1588
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Re: Mymensingh 2008 /5
I think likewise, The greatest number of primes shud be in the interval 51 to 71, However i m having difficulty proving it
Bristy Sikder
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Abdul Muntakim Rafi
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Re: Mymensingh 2008 /5
Let the 20 consecutive numbers be
$n,n+1,n+2,....................,n+19$
10 of them will be divisible by 2.
We have 10 numbers left.
Let n be an odd number...
So the numbers which remain left are
$n,n+2,.................n+19$
We know the multiples of are like this...
$3k,3k+1,3k+2,3(k+1),.................$
so n can be presented as $3k,3k+1,3k+2$ any one of the three.
Just argue with this... and we will see at least three numbers of this ten numbers will be divisible by 3.
As the problem demands the greatest possibility of most numbers being prime, we just take n to be 3k+1 or 3k+2. If we take n to be 3k that'd lead to 4 non primes among these 10 numbers...
Now we got 7 numbers left... At least 2 of them will be divided by 5... If n is divisible by 5 then 3 will be divisible by 5.But we can skip that.
So we got 5 left...
No matter what happens 15 numbers will be divisible by 2,3,5 in any 20 consecutive numbers.
Now whatever happens we can't get more than 5 primes... We can get less than 5... But we got 5 primes from
$51-70$... So that'd lead us to a conclusion that among 20 consecutive integers 5 numbers can be prime...
$n,n+1,n+2,....................,n+19$
10 of them will be divisible by 2.
We have 10 numbers left.
Let n be an odd number...
So the numbers which remain left are
$n,n+2,.................n+19$
We know the multiples of are like this...
$3k,3k+1,3k+2,3(k+1),.................$
so n can be presented as $3k,3k+1,3k+2$ any one of the three.
Just argue with this... and we will see at least three numbers of this ten numbers will be divisible by 3.
As the problem demands the greatest possibility of most numbers being prime, we just take n to be 3k+1 or 3k+2. If we take n to be 3k that'd lead to 4 non primes among these 10 numbers...
Now we got 7 numbers left... At least 2 of them will be divided by 5... If n is divisible by 5 then 3 will be divisible by 5.But we can skip that.
So we got 5 left...
No matter what happens 15 numbers will be divisible by 2,3,5 in any 20 consecutive numbers.
Now whatever happens we can't get more than 5 primes... We can get less than 5... But we got 5 primes from
$51-70$... So that'd lead us to a conclusion that among 20 consecutive integers 5 numbers can be prime...
Man himself is the master of his fate...