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m\\)<\/span>.\n(This follows from the pigeonhole principle and the fact that\nthere's a finite number of permutations of the Rubik's cube.)\nIf <span class=\"mathjax mathjax--inline\">\\(M^{-1}\\)<\/span> denotes the inverse permutation, then<\/p>\n<p class=\"mathjax mathjax--block\">\\[\nM^m = M^n \\iff M^{m-n} = M^0 ~ .\\]<\/p>\n<p>This shows that <span class=\"mathjax mathjax--inline\">\\(M^{m - n}\\)<\/span> (where <span class=\"mathjax mathjax--inline\">\\(m - n \\geq 0\\)<\/span>) is the solved cube.<\/p>\n<h3 id=\"overview-of-less-math-y-solution\">Overview of less math-y solution<a href=\"#overview-of-less-math-y-solution\" class=\"toc-anchor after\" data-anchor-icon=\"#\" aria-label=\"Anchor\"><\/a><\/h3>\n<p>Here is the generic overview of the solution we will produce together:<\/p>\n<ol><li>we prove that if we keep scrambling the cube, we...<\/li><\/ol>","summary":"This problem explores what happens when you try to scramble a Rubik&#039;s cube systematically.","date_modified":"2025-07-23T16:49:02+02:00","tags":["combinatorics","group theory","logic","mathematics"],"image":"\/user\/pages\/02.blog\/03.problems\/p043-rubiks-cube-scrambling\/thumbnail.png"}]}
