Problem 050
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Problem 050
I think that something's going wrong! The number 953 is prime but it isn't the longest sum of consecutive primes below onethousand that adds to a prime. The number 281 has this property with 14 terms.
According to this, the prime, below onemillion, which can be written as the sum of the most consecutive primes, is 958577 with 536 terms, but my answer is wrong. I wonder why?
According to this, the prime, below onemillion, which can be written as the sum of the most consecutive primes, is 958577 with 536 terms, but my answer is wrong. I wonder why?
Re: incorect example in problem 50?
The problem states that 953 is a sum of 21 consecutive primes (actually : 7+11+13+...+83+89).
Isn't that a longer sum than the 14 terms adding up to 281 ?
Isn't that a longer sum than the 14 terms adding up to 281 ?
Re: incorect example in problem 50?
ok thanks!
I thouhgt that the sum should begin always from 2 (the first prime)
I thouhgt that the sum should begin always from 2 (the first prime)
Problem 050
These consecutive numbers start from 2 or any prime number?
Re: Problem 050
Well, as you can see there was already a topic for this problem (I have just merged the two topics). So:
(a) There was no need to start a new topic.
(b) The answer to your question was there already
(a) There was no need to start a new topic.
(b) The answer to your question was there already
Re: Problem 050
Thanks harryh  I can swear I searched for "Problem 050" but nothing showed up on the search results  Do they get archived after a period of neglect?
 daniel.is.fischer
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 Joined: Sun Sep 02, 2007 11:15 pm
 Location: Bremen, Germany
Re: Problem 050
No, but the search term "Problem xxx" is too generic to produce results.
Go to the bottom of the forum page, there you have the opportunity to change by which criteria the topics are sorted. Sort by subject, and it's easy to find (if it's there already).
Go to the bottom of the forum page, there you have the opportunity to change by which criteria the topics are sorted. Sort by subject, and it's easy to find (if it's there already).
Il faut respecter la montagne  c'est pourquoi les gypaètes sont là.
Re: Problem 050
The problem states:
"The longest sum of consecutive primes below onethousand that adds to a prime, contains 21 terms, and is equal to 953."
II always come up with something different. I get 961 as a sum of 23 primes.
961 = 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 + 43 + 47 + 53 + 59 + 61 + 67 + 71 + 73 + 79 + 83 + 89
Can anyone verify if my answer is incorrect? I have solved the problem, but I'm curious about this discrepancy.
"The longest sum of consecutive primes below onethousand that adds to a prime, contains 21 terms, and is equal to 953."
II always come up with something different. I get 961 as a sum of 23 primes.
961 = 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 + 43 + 47 + 53 + 59 + 61 + 67 + 71 + 73 + 79 + 83 + 89
Can anyone verify if my answer is incorrect? I have solved the problem, but I'm curious about this discrepancy.

 Posts: 53
 Joined: Mon Oct 20, 2008 10:48 pm
Re: Problem 050
961=31*31, so that sum is not a prime.srmackey wrote:The problem states:
"The longest sum of consecutive primes below onethousand that adds to a prime, contains 21 terms, and is equal to 953."
II always come up with something different. I get 961 as a sum of 23 primes.
961 = 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 + 43 + 47 + 53 + 59 + 61 + 67 + 71 + 73 + 79 + 83 + 89
Can anyone verify if my answer is incorrect? I have solved the problem, but I'm curious about this discrepancy.
Re: Problem 050
Ah... good call. There must be something wrong with my sieve algorithm. Thanks for the feedback!genious999 wrote:961=31*31, so that sum is not a prime.
Re: Problem 050
Why the largest below 1000 is
953 = 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 + 43 + 47 + 53 + 59 + 61 + 67 + 71 + 73 + 79 + 83 + 89
and not
983 = 23 + 29 + 31 + 37 + 41 + 43 + 47 + 53 + 59 + 61 + 67 + 71 + 73 + 79 + 83 + 89 + 97
Both are primes and the sum of primes in sequence.
So the largest should be 983. What I'm doing wrong here?
Ok I got it... is the largest in number of terms.
953 = 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 + 43 + 47 + 53 + 59 + 61 + 67 + 71 + 73 + 79 + 83 + 89
and not
983 = 23 + 29 + 31 + 37 + 41 + 43 + 47 + 53 + 59 + 61 + 67 + 71 + 73 + 79 + 83 + 89 + 97
Both are primes and the sum of primes in sequence.
So the largest should be 983. What I'm doing wrong here?
Ok I got it... is the largest in number of terms.
Re: Problem 050
Why did you skip 2,3, and 5? At 21 numbers, I get 712, but 281 was the largest so I'm wrong somewhere.abranches wrote:Why the largest below 1000 is
953 = 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 + 43 + 47 + 53 + 59 + 61 + 67 + 71 + 73 + 79 + 83 + 89
and not
983 = 23 + 29 + 31 + 37 + 41 + 43 + 47 + 53 + 59 + 61 + 67 + 71 + 73 + 79 + 83 + 89 + 97
Both are primes and the sum of primes in sequence.
So the largest should be 983. What I'm doing wrong here?
Ok I got it... is the largest in number of terms.
Re: Problem 050
712 is not prime.JMW1994 wrote:Why did you skip 2,3, and 5? At 21 numbers, I get 712, but 281 was the largest so I'm wrong somewhere.
Re: Problem 050
Well, I knew that. The problem states:thundre wrote:712 is not prime.JMW1994 wrote:Why did you skip 2,3, and 5? At 21 numbers, I get 712, but 281 was the largest so I'm wrong somewhere.
When I tried to solve the largest onethousand consecutive prime and I had 21 prime terms,including 2, 3, and 5, I got the number 712 which is obviously a composite number. I got 281 as the largest which wasn't correct. Besides, I never got and answer from my last question. The question was: Why did you skip 2,3, and 5?The prime 41, can be written as the sum of six consecutive primes:
41 = 2 + 3 + 5 + 7 + 11 + 13
This is the longest sum of consecutive primes that adds to a prime below onehundred.
The longest sum of consecutive primes below onethousand that adds to a prime, contains 21 terms, and is equal to 953.
Which prime, below onemillion, can be written as the sum of the most consecutive primes?
Re: Problem 050
Why not? The question states only that the sequence of primes is consecutive, not that it must start at 2.JMW1994 wrote:The question was: Why did you skip 2,3, and 5?
_{Jaap's Puzzle Page}
Re: Problem 050
My next question is why isn't 2, 3, or 5 consecutive?
Re: Problem 050
Your questions aren't really making much sense.. 2, 3, and 5 are consecutive primes; they're even shown in the example in the problem as being consecutive primes. Nobody has said that they aren't.
Project 50
I have a question about the directions for this problem. When you click on the problem it gives the following examples:
it adds the first 6 primes and gets 41 then says that this: This is the longest sum of consecutive primes that adds to a prime below onehundred. I get this.
Then it says:
The longest sum of consecutive primes below onethousand that adds to a prime, contains 21 terms, and is equal to 953.
But when I add the first 21 prime numbers together, I get 712, not 953. When I add the first 24 primes together, I get 963, completely skipping 953. This makes me think that I don't quite understand what the problem is asking.
Can someone explain how the first 21 terms gives 953?
Thanks.
it adds the first 6 primes and gets 41 then says that this: This is the longest sum of consecutive primes that adds to a prime below onehundred. I get this.
Then it says:
The longest sum of consecutive primes below onethousand that adds to a prime, contains 21 terms, and is equal to 953.
But when I add the first 21 prime numbers together, I get 712, not 953. When I add the first 24 primes together, I get 963, completely skipping 953. This makes me think that I don't quite understand what the problem is asking.
Can someone explain how the first 21 terms gives 953?
Thanks.
Re: Project 50
The sum does not have to start with 2.
e.g. 7+11+13 = 31 would be another example.
e.g. 7+11+13 = 31 would be another example.

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 Location: California, USA
Re: Project 50
When the problem says
"The longest sum of consecutive primes below onethousand that adds to a prime, contains 21 terms, and is equal to 953."
That does not mean
from 2 to the 21st prime number below onethousand.
Whenn you add X amonunt of consecutive primes below that upperbound, it should result in being the longest chain within onthousand.
Good luck.
"The longest sum of consecutive primes below onethousand that adds to a prime, contains 21 terms, and is equal to 953."
That does not mean
from 2 to the 21st prime number below onethousand.
Whenn you add X amonunt of consecutive primes below that upperbound, it should result in being the longest chain within onthousand.
Good luck.