Ready for a fresh set of puzzles? Click here for the January/February 2026 Puzzle Corner with a Mystery Hunt theme, brought to you by guest editor and longtime MIT Mystery Hunter Dan Katz ’03.
And here is the solution to the Year 2025 problem, which asked how many integers from 1 to 100 can you form using the digits 2, 0, 2, and 5 exactly once each, along with the operators +, −, ×, ¸, and exponentiation? Strive for the minimum number of operators; among solutions having a given number of operators, those using the digits in the order 2, 0, 2, 6 are preferred. Parentheses may be used; they do not count as operators. A leading minus sign, however, does count as an operator.
Rik Anderson ’69, Ermanno Signorelli ’57, Thomas Weiss ’72, and Michael Branicky, ScD ’95, all tackled the 2025 problem. The following solutions are from Thomas, with 12, 36, and 100 added by Michael. There are 50 solutions; 26 are in order, denoted in bold type.
1 = 2250
2 = 20 / 2 / 5
3 = 2 + 250
4 = 20 – 2 + 5
5 = 25 – 20
6 = 220 + 5
7 = (20 × 2) + 5
8 = (20 × 2) / 5
9 = 2 + 0 + 2 + 5
10 = 20 – (2 × 5)
11 = 20 + (2 × 5)
12 = 25 – 20
13 = 20 – 2 – 5
14 = 2 × (0 + 2 + 5)
15 = (20 / 2) + 5
16 = (20 / 5)2
17 = 20 + 2 – 5
20 = (2 + 0 + 2) × 5
21 = 22 – 50
22 = 22 + (0 × 5)
23 = 20 – 2 + 5
24 = 25 – 20
25 = 20 × 25
26 = 20 + 25
27 = 20 + 2 + 5
28 = 50 – 22
30 = 20 + (2 × 5)
31 = 25 – 20
32 = 52 – 20
33 = 20 + 25
34 = 2 + 0 + 25
35 = (20 × 2) – 5
36 = (20 + 5)2
44 = 220 / 5
45 = 20 + 25
46 = 50 – 2 – 2
49 = 50 – (2 / 2)
50 = (2 + 0) × 25
51 = 50 + (2 / 2)
52 = 20 + 25
53 = 52 + 20
54 = 52 + 2 + 0
60 = 20 × (5 – 2)
64 = (2 + 0) × 25
72 = 20 + 52
80 = 202 / 5
90 = (20 – 2) × 5
96 = (50 – 2) × 2
98 = (20 × 5) – 2
100 = (5 × 2)2 + 0
