HMMT 十一月 2025 · THM 赛 · 第 7 题
HMMT November 2025 — THM Round — Problem 7
题目详情
- Io, Europa, and Ganymede are three of Jupiter’s moons. In one Jupiter month, they complete exactly I , E , and G orbits around Jupiter, respectively, for some positive integers I , E , and G . Each moon appears as a full moon precisely at the start of each of its orbits. Suppose that in every Jupiter month, there are • exactly 54 moments of time with at least one full moon, • exactly 11 moments of time with at least two full moons, and • at least 1 moment of time with all three full moons. Compute I · E · G .
解析
- Io, Europa, and Ganymede are three of Jupiter’s moons. In one Jupiter month, they complete exactly I , E , and G orbits around Jupiter, respectively, for some positive integers I , E , and G . Each moon appears as a full moon precisely at the start of each of its orbits. Suppose that in every Jupiter month, there are • exactly 54 moments of time with at least one full moon, • exactly 11 moments of time with at least two full moons, and • at least 1 moment of time with all three full moons. Compute I · E · G . Proposed by: Derek Liu, Eric Wang Answer: 7350 Solution: Note that if two moons appear as full moons exactly m and n times each month, respectively, then they are simultaneously full moons exactly gcd( m, n ) times each month. Using inclusion-exclusion, we can interpret the problem conditions as I + E + G − gcd( I, E ) − gcd( E, G ) − gcd( I, G ) + gcd( I, E, G ) = 54 , gcd( I, E ) + gcd( E, G ) + gcd( I, G ) − 2 gcd( I, E, G ) = 11 . From this, we see that gcd( I, E, G ) divides both 54 and 11, so gcd( I, E, G ) = 1. Thus, gcd( I, E ) + gcd( E, G ) + gcd( I, G ) = 11 + 2 = 13 and I + E + G = 54 + 13 − 1 = 66 . Now, consider the values of gcd( I, E ), gcd( E, G ), and gcd( I, G ). They must add to 13, and the gcd of any two of them is gcd( I, E, G ) = 1. In particular, since their sum is 13, either 0 or 2 of them are even, so none of them are even. It follows that the only possibilities for the three gcds are (1 , 1 , 11) and (1 , 5 , 7). Note that if 11 divides two of I , E , and G , it must divide the third, since it divides their sum I + E + G = 66. Hence, (1 , 1 , 11) is not possible, leaving (1 , 5 , 7) as the only possibility. Thus, I , E , and G must be of the form 35 a , 5 b , and 7 c , in some order, for some positive integers a , b , and c . The only solution satisfying I + E + G = 66 is ( a, b, c ) = (1 , 2 , 3), resulting in I , E and G being 35, 10, and 21, in some order. It can be checked that these values satisfy the given conditions, so I · G · E = 35 · 10 · 21 = 7350 .