## Problem 050

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persa
Posts: 2
Joined: Mon Jul 28, 2008 2:00 pm

### 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 one-thousand that adds to a prime. The number 281 has this property with 14 terms.
According to this, the prime, below one-million, 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?

harryh
Posts: 2091
Joined: Tue Aug 22, 2006 9:33 pm
Location: Thessaloniki, Greece

### 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 ?

persa
Posts: 2
Joined: Mon Jul 28, 2008 2:00 pm

### Re: incorect example in problem 50?

ok thanks!
I thouhgt that the sum should begin always from 2 (the first prime)

iamhigh
Posts: 7
Joined: Fri Apr 03, 2009 11:24 am

### Problem 050

These consecutive numbers start from 2 or any prime number?

harryh
Posts: 2091
Joined: Tue Aug 22, 2006 9:33 pm
Location: Thessaloniki, Greece

### 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 iamhigh
Posts: 7
Joined: Fri Apr 03, 2009 11:24 am

### 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
Posts: 2400
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).
Il faut respecter la montagne -- c'est pourquoi les gypa&egrave;tes sont l&agrave;.

srmackey
Posts: 2
Joined: Wed Nov 04, 2009 1:55 pm

### Re: Problem 050

The problem states:

"The longest sum of consecutive primes below one-thousand 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

genious999
Posts: 53
Joined: Mon Oct 20, 2008 10:48 pm

### Re: Problem 050

srmackey wrote:The problem states:

"The longest sum of consecutive primes below one-thousand 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

961=31*31, so that sum is not a prime.

srmackey
Posts: 2
Joined: Wed Nov 04, 2009 1:55 pm

### Re: Problem 050

genious999 wrote:961=31*31, so that sum is not a prime.
Ah... good call. There must be something wrong with my sieve algorithm. Thanks for the feedback!

abranches
Posts: 1
Joined: Sat Nov 20, 2010 2:36 pm

### 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.

JMW1994
Posts: 43
Joined: Sat Apr 09, 2011 11:35 pm

### Re: Problem 050

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.
Why did you skip 2,3, and 5? At 21 numbers, I get 712, but 281 was the largest so I'm wrong somewhere. thundre
Posts: 356
Joined: Sun Mar 27, 2011 10:01 am

### Re: Problem 050

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.
712 is not prime. JMW1994
Posts: 43
Joined: Sat Apr 09, 2011 11:35 pm

### Re: Problem 050

thundre wrote:
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.
712 is not prime.
Well, I knew that. The problem states:
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 one-hundred.

The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.

Which prime, below one-million, can be written as the sum of the most consecutive primes?
When I tried to solve the largest one-thousand 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? jaap
Posts: 551
Joined: Tue Mar 25, 2008 3:57 pm
Contact:

### Re: Problem 050

JMW1994 wrote:The question was: Why did you skip 2,3, and 5?
Why not? The question states only that the sequence of primes is consecutive, not that it must start at 2.
Jaap's Puzzle Page JMW1994
Posts: 43
Joined: Sat Apr 09, 2011 11:35 pm

### Re: Problem 050

My next question is why isn't 2, 3, or 5 consecutive? TripleM
Posts: 382
Joined: Fri Sep 12, 2008 3:31 am

### 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.

apdwyer51
Posts: 5
Joined: Sun Nov 27, 2011 8:46 pm

### 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 one-hundred. I get this.

Then it says:

The longest sum of consecutive primes below one-thousand 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.

wrongrook
Posts: 397
Joined: Sat Oct 17, 2009 10:39 pm

### Re: Project 50

e.g. 7+11+13 = 31 would be another example.

Eternalcode
Posts: 12
Joined: Mon Oct 03, 2011 2:25 am
Location: California, USA

### Re: Project 50

When the problem says

"The longest sum of consecutive primes below one-thousand 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 one-thousand.

Whenn you add X amonunt of consecutive primes below that upperbound, it should result in being the longest chain within on-thousand.

Good luck. 