How can the year 2011 be expressed as the sum of consecutive primes in MATLAB 7.11 (R2010b) ?

I heard that the number 2011 can be expressed as the sum of 11 consecutive prime numbers. How can I find those numbers in MATLAB?

Answer 1

[Walter Roberson]

t = 1:2011;t = t(isprime(t)); p = cumsum(t);t(find(p(12:end) - p(1:end-11) == 2011) + (1:11))

Answer 2

[Matt Fig]

Another method:

N = 2011;  % Given
P = primes(N/4);  % Conservative estimate for ceiling
M = N/11;  % The mean
[I,I] = min(abs(filter(ones(1,11)/11,1,P)-M)); % Moving average comparison.
SOL = P(I-10:I) % The desired primes.
sum(SOL)==N  %Check that the solution is good

Answer 3

[John D'Errico]

There are many solutions to this. Brute force is one.

>> p = primes(2011);
>> find(conv(p,ones(1,2)) == 2011)
ans =
   306
>> find(conv(p,ones(1,3)) == 2011)
ans =
   123   307
>> find(conv(p,ones(1,4)) == 2011)
ans =
   308
>> find(conv(p,ones(1,5)) == 2011)
ans =
   309
>> find(conv(p,ones(1,6)) == 2011)
ans =
   310
>> find(conv(p,ones(1,7)) == 2011)
ans =
   311
>> find(conv(p,ones(1,8)) == 2011)
ans =
   312
>> find(conv(p,ones(1,9)) == 2011)
ans =
   313
>> find(conv(p,ones(1,10)) == 2011)
ans =
   314
>> find(conv(p,ones(1,11)) == 2011)
ans =
    47   315

See that there are two cases where find found more than 1 result that yielded 2011. They correspond to the case for a sum of 3 consecutive primes, and 11 consecutive primes.

>> p(123 + [-2:0])
ans =
   661   673   677
>> sum(ans)
ans =
        2011

Here is the set of 11 primes.

>> p(47 + [-10:0])
ans =
   157   163   167   173   179   181   191   193   197   199   211
>> sum(ans)
ans =
        2011

However, you can be more creative. If a set of 11 primes will sum to 2011, then they must average...

>> 2011/11
ans =
       182.82

Therefore the middle prime, if a set of 11 consecutive primes will solve this problem, must be roughly 183. The closest prime to 183 is 181. Testing the set of 5 primes below that value and the 5 above it does yield 2011, as we showed above. We can use similar logic to find a set of 17 consecutive primes that sum to the number 17717.

>> 17717/17
ans =
       1042.2

Checking the 17 primes that bracket 1042 or so, we find this set, which does sum to the value I specified.

>> sum([991  997  1009  1013  1019  1021  1031  1033  1039  1049  1051  1061  1063  1069  1087  1091  1093])
ans =
       17717

[Matt Fig]

If you notice, that last method is what I was doing ;-) I am wondering if M*2 would be a better initial estimate.

[John D'Errico]

Yes, it is. An interesting problem is to search for numbers which are a sum of consecutive primes in more than one way. For example, what is the smallest number that is the sum of consecutive primes in at least 2 ways? In 3 distinct ways?

[John D'Errico]

For example, 36 is representable as the sum of 2 and 4 consecutive primes.

>> sum([5 7 11 13])

ans =

36

>> sum([17 19])

ans =

36

Likewise, 41 can be written as the sum of both 3 and 6 consecutive primes. But what is the smallest number that is representable as such a sum in 3 distinct ways?

[Matt Fig]

I get 1151 for the smallest such number.

[449 457 461]

[131 137 139 149 151 157 163 167 173]

[101 103 107 109 113 127 131 137 139 149 151]

[John D'Errico]

Are you sure?

>> 449+457+461

ans =

1367

>> sum([101 103 107 109 113 127 131 137 139 149 151])

ans =

1367

I found 240 as the smallest number with 3 distinct sums, then 311 as the smallest representable as 4 distinct sums of primes.

>> sum([101 103 107])

ans =

311

>> sum([53 59 61 67 71])

ans =

311

>> sum([31 37 41 43 47 53 59])

ans =

311

>> sum([11 13 17 19 23 29 31 37 41 43 47])

ans =

311

[Matt Fig]

A yes. I was only considering primes, as 2011 is prime. I copied the results for the second prime who has 3 unique sums by mistake. Here are the numbers for 1151:

[379 383 389]

[223 227 229 233 239]

[7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101]

I guess I misunderstood what you were after.

Answer 4

[Doug Hull]

stack = [];

for i = 1:2011
  if isprime(i)
      stack = [i stack];
  end
  if numel(stack) > 11
      stack = stack(1:11);
  end
  if sum(stack) == 2011
      disp(sum(stack))
      disp(stack)
      break
  end
end