This is New Years, not Prime Day!

The number $2025$ is not prime. As a matter of fact, it’s a perfect square: $2025 = 45^2$.

You cannot make $2025$ by adding two distinct primes. To do so, you’d have to add an even prime and an odd prime. The only even prime is $2,$ but $2025 − 2 = 2023$, which is not prime (it’s equal to $7∙172$). But you can make $2025$ by adding three distinct primes. For example, $661 + 673 + 691 = 2025.$ You can also make $2025$ by adding four distinct primes: $2 + 659 + 673 + 691 = 2025.$

What is the greatest number of distinct primes that add up to $2025$?

First let's find the set of all primes less than $2025,$ that is, \begin{align*}P = \{ & 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101,\\ & 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197,\\ & 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311,\\ & 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431,\\ & 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557,\\ & 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661,\\ & 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809,\\ & 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937,\\ & 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049,\\ & 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153,\\ & 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277,\\ & 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381,\\ & 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487,\\ & 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597,\\ & 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699,\\ & 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823,\\ & 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949,\\ & 1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003, 2011, 2017 \}.\end{align*} One way to encode the problem is as the following binary linear knapsack program: \begin{align*} N = \max \,\, & \sum_{p \in P} x_p \\ \text{s.t.} \,\, & \sum_{p \in P} p x_p = 2025 \\ & x_p \in \{ 0, 1 \}, \, \forall p \in P\end{align*}

We can first loosen the binary restrictions a bit to get an upper bound. If as opposed to having $x_p \in \{0,1\},$ if we use $x_p \in [0,1],$ $\forall p \in P,$ then we get an upper bound of $\tilde{N} = \frac{4624}{139} = 33.266187\dots,$ by greedily filling our knapsack with all of the smallest primes until we have to resort to a fractional piece, that is, $$\tilde{x}_p = \max \left\{ 0, \min \left\{ \frac{2025 - \sum_{j \in P, j \lt p } j }{p}, 1 \right\} \right\}, \, \, \forall p \in P.$$ So we know that $N \leq \lfloor \tilde{N} \rfloor = 33.$ If we have $33$ primes and $2$ is one of them, then the sum will be even and so could never be $2025.$ The smallest sum of $33$ primes that does not include $2$ is $2125,$ which is too large. So we can never have a set of $33$ primes add up to $2025,$ and hence the upper bound must be $N \leq 32.$ From here, we can either hunt and peck for the right tradeoff of primes, ... or we can just program a quick binary program and your local solver to arrive at the optimal answer is that you can make $2025$ into the sum of $N = 32$ primes (so not that far off from the non-integral approximation), since \begin{align*}2025 &= 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 + 47 \\ & + 53 + 59 + 61 + 67 + 71 + 73 + 79 + 83 + 89 + 101 + 103 + 107\\ & + 109 + 113 + 127 + 139 + 149 + 157 \end{align*}

It's my particles in a box!

You have three particles inside a unit square that all repel one another. The energy between each pair of particles is $1/r,$ where $r$ is the distance between them. To be clear, the particles can be anywhere inside the square or on its perimeter. The total energy of the system is the sum of the three pairwise energies among the particles.

What is the minimum energy of this system, and what arrangement of the particles produces it?

If we only had two particles, then the furthest away that the particles could be (and hence the minimal total energy arrangment), would be opposite corners of the unit square, e.g., one at $(0,0)$ and the other at $(1,1).$ In this configuration, the total energy would be $E_2 = \frac{1}{\sqrt{2}}.$ So let's assume that we have two of the three particles there and then see where the other particle would end up.

Assume that the third particle is at the point $X=X(a,b)$ for some $0 \leq a, b \leq 1.$ Then the total energy of the system would be $$E = E(a,b) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{a^2+b^2}} + \frac{1}{\sqrt{(1-a)^2 + (1-b)^2}}.$$ Taking derivatives gives $$\nabla E = \begin{pmatrix} -\frac{a}{(a^2 + b^2)^{3/2}} + \frac{1-a}{\left( (1-a)^2 + (1-b)^2 \right)^{3/2}} \\ -\frac{b}{(a^2 + b^2)^{3/2}} + \frac{1-b}{\left( (1-a)^2 + (1-b)^2 \right)^{3/2}}\end{pmatrix}.$$ The only analytic root of the gradient mapping is $a=b=\frac{1}{2},$ which would give a total energy of $E(\frac{1}{2},\frac{1}{2}) = \frac{1}{\sqrt{2}} + 2 \frac{2}{\sqrt{2}} = \frac{5}{\sqrt{2}},$ but this is a saddle point and we can do better by checking what happens if $X$ were along the perimeter of the unit square.

Assume without loss of generality and by availing ourselves of symmetry, that $b = 0$ and we just have $0 \leq a \leq 1.$ In this case, we have $$\tilde{E}(a) = E(a,0) = \frac{1}{\sqrt{2}} + \frac{1}{a} + \frac{1}{\sqrt{1 + (1-a)^2}}$$ which has derivative $$\tilde{E}^\prime = -\frac{1}{a^2} + \frac{1-a}{\left(1 + (1-a)^2\right)^{3/2}} = \frac{a^2(1-a) - \left(a^2 - 2a+2\right)^{3/2}}{a^2\left(a^2 - 2a+2\right)^{3/2}}.$$ Since $$a^2(1-a) \leq \frac{4}{27} \lt 1 \leq \left(a^2 - 2a + 2\right)^{3/2},$$ for all $a \gt 0,$ we have $\tilde{E}^\prime \lt 0$ for all $a \gt 0,$ so the lowest possible value is when $a^* = 1,$ that is when $X^* = (1,0),$ and the total energy is at its lowest $$E^* = \tilde{E}(1) = \frac{1}{\sqrt{2}} + 2 \approx 2.70710678119....$$ There are other possible arrangements of the particles that obtain the same minimal energy. Namely, any of the four right isoceles triangle formed by choosing three out of the four vertices of the unit square will give this minimal energy configuration.