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 $$\begin{align*}2025 &= 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 + 47 \\ & + 53 + 59 + 61 + 67 + 71 + 73 + 79 + 83 + 89 + 101 + 103 + 107\\ & + 109 + 113 + 127 + 139 + 149 + 157 \end{align*}$$