parmenides51 wrote: Maria and Luis play the following game: Maria throws three dice and Luis can select some of them (possibly none) and turn them changing their value for the value in the opposite face of each selected die. Prove that Luis can always play in such a way that the sum of the upper faces of the dice after the change is a multiple of $4$. Note: The game is played with normal dice, that is, the sum of opposite faces is $7$. seeing in mod 4 the numbers of the die are 1, 2, -1, 0, 1 and 2, also as the sum of opposite faces must add 7 we can change the 1 by 2, the 2 by 1, the -1 by 0 and vice versa, then the possible triples of numbers that appear on the upper faces in mod 4 are: (1, 2, -1); (1, 2, 0); (2, 2, -1); (2, 2, 0); (2, 1, -1); (2, 1, 0); (1, 1, -1); (1, 1, 0) in the first triple if we change the 1 by 2, and the -1 by 0 and the sum of the 3 numbers is 0 mod 4. in the second triple if we change the 1 by 2, and the sum of the 3 numbers is 0 in mod 4. In the third triple, if we change the -1 by 0, the sum of the 3 numbers is 0 in mod 4. in the fourth triple if we do not need to change anything, because the sum is already 0 in mod 4. In the fifth triple we change the 1 to 2, and the -1 to 0, and the sum of the 3 numbers is already 0 in mod 4. In the sixth triple we change the 1 to 2, and the sum is already 0 in mod 4. In the seventh triple we change the two 1s for 2 two, and the -1 for 0, and the sum of the 3 is 0 in mod 4. In the eighth triple, change the two 1s by 2 two, and the sum is already a multiple of 4. finally we see that in any case it is possible to change the numbers such that the sum of the numbers on the faces is a multiple of 4.