2025 puzzle

This week’s Fiddler is about the number 2025, in celebration of (almost) New Years!

First puzzle: What is the greatest number of distinct primes that add up to 2025?

Second puzzle: How can you assign a set of 20 distinct prime numbers to the 20 vertices of a dodecahedron, so that the numbers on the five vertices of each face add up to 2025?

My solution:
[Show Solution]

4 thoughts on “2025 puzzle”

  1. Very nice work on a difficult 2nd problem. I took a look at the problem over the weekend and gave up, leading me to 3 questions. First, in choosing a diagram for display purposes, why go with the dodec. “net” instead of a platonic graph (also called a Schlegel graph)? The graph has 20 nodes for the 20 vertices, making the prime assignments a little more direct, I would think. Second, didn’t you previously analyze this question, or something similar, in February 2022, back when the Fiddler was the Riddler? That question was about assigning distinct primes to the vertices of cube so that each face had the same sum. The difference there, as I recall, was that you could choose the sum, rather than being forced to hit a target number. Which brings me to my third question. How much of your prior analysis in 2022 could still be applied to the dodecahedron primes with a target sum? Nothing against optimization solvers, but I am particularly interested in how to break down problems like this analytically, to the extent possible, before we all run to our computers.

    1. Thank you for the comment! I had totally forgotten about that old Riddler problem… I updated my solution with a link to it and also a Schlegel diagram of the solution.

      Regarding your question, I think that having a variable sum was a critical feature that made the previous Riddler problem amenable to an analytic approach. Namely, it relied on the Green-Tao Theorem on the existence of arithmetic progressions of primes. Forcing the primes to lie in arithmetic progressions makes it easy to find solutions, but they end up having very large sums… If we want the sums to be smaller, such as 2025, I think that we can no longer expect to find solutions that lie in arithmetic progressions.

      In general, I think this is a challenging question because there are very few known results about the structure of prime numbers. Beyond arithmetic progressions, I’m not sure what sort of structure we can exploit.

  2. In the spriit of Eugene Wigner (who allegedly once said “It’s nice to know that the computer understands the problem, but I would like to understand it too”), here is how we can see that the max number of primes that sum to 2025 is 32. First, add up the smallest primes until the sum exceeds 2025: this happens with 34 primes, with a sum of 2127. So the max number that sums to 2025 must be less than 34. Could it be 33? Only if 2 is excluded, otherwise the sum is even. But the sum of the 33 smallest odd primes is just the previous sum minus 2, or 2125: still too big. So the largest possible number of primes that sums to 2015 must be 32 or fewer. And then there are many solutions with 32 primes, including 56 in which we add the 35th prime (149) and drop three other smaller primes. Your solution uses primes up to the 43rd (191).

    Happy holidays and especially Happy New Year!

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