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jasperE3 wrote: Given $8$ unit cubes, $24$ of their faces are painted in blue and the remaining $24$ faces in red. Show that it is always possible to assemble these cubes into a cube of edge $2$ on whose surface there are equally many blue and red unit squares. See here page 3 in croatian. Translation by chatGPT: Original statement: We have 8 cubes with edge length 1, 24 faces of which are painted blue, and the remaining 24 red. Prove that from these cubes, it is possible to form a cube (2 × 2 × 2) whose surface will have an equal number of blue and red squares (1 × 1). Sol: Let's consider an arbitrary \(2 \times 2 \times 2\) cube, and let its surface have \(m\) blue squares and \(n\) white squares, with \(m \approx n\). If \(m = n\), then we are done. Otherwise, let \(d = m - n > 0\). Here, \(d\) is an even number because \(m + n = 24\), so \[ d = m - n = 24 - 2n = 2(12 - n). \]Let's consider the following moves. By rotating the cube around one of its axes by 90°, exactly one side becomes hidden while another side becomes visible. In this process, \(d\) either remains the same or changes by 2. Using three such rotations, we can turn the cube so that all three visible sides become hidden and vice versa. If we do this with all the cubes, the visible sides will become all the hidden sides, resulting in \(n\) blue and \(m\) white squares. Therefore, the new \(d' = n - m = -d\). Hence, as \(d\) changes by 2 or 0 with successive rotations, and since we have achieved that \(d\) changes sign, and \(d\) is even, this means that at some point \(d\) must be equal to zero, which is what we needed to prove.