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Solution or solutions (Posted on 2014-03-20) Difficulty: 3 of 5
The sum of n consecutive primes beginning
with n is n times the nth prime.

1. I know one solution - find it!
2. Are there any others? - please comment.

No Solution Yet Submitted by Ady TZIDON    
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Some Thoughts computer evaluation | Comment 2 of 6 |

  10    while Ct<44
  20      P0=nxtprm(P0) ' p0 is n
  30      Sum=P0:P1=P0
  40      for I=2 to P0
  50        P1=nxtprm(P1)
  60        Sum=Sum+P1
  70      next
  80      P2=prm(P0)
  90      Prod=P0*P2
  95      if Sum=Prod then
 100      :print P0,Sum;Prod
 102      :print P0;:P1=P0:for I=2 to P0:P1=nxtprm(P1):print P1;:next
 103      :print:print
 110      :inc Ct
 120    wend
 
stops while attempting to find the 12,253th prime, but up to that time, it finds only two solutions:

 3       15  15
 3  5  7

 7       119  119
 7  11  13  17  19  23  29
 
The sum of 3 consecutive primes starting with 3, and of 7 consecutive primes starting with 7 are respectively equal to the 3 times the third prime (5) and 7 times the 7th prime (17).
 
The respective sets of 3 and of 7 primes are shown below the initial prime and the sum involved and the (equal) product involved.


  Posted by Charlie on 2014-03-20 15:34:49
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