{"id":47,"date":"2017-04-01T03:10:04","date_gmt":"2017-03-31T19:10:04","guid":{"rendered":"http:\/\/colliot.me\/?p=47"},"modified":"2017-11-20T07:12:31","modified_gmt":"2017-11-19T23:12:31","slug":"a-simple-combinatorial-problem-and-related-thoughts","status":"publish","type":"post","link":"https:\/\/colliot.org\/zh\/2017\/04\/a-simple-combinatorial-problem-and-related-thoughts\/","title":{"rendered":"\u4e00\u4e2a\u7b80\u5355\u7684\u7ec4\u5408\u95ee\u9898\u53ca\u76f8\u5173\u7684\u601d\u8003"},"content":{"rendered":"

Problem:<\/b> If the decimal expansion of a<\/span> contains all 10<\/span> digits (0,...,9<\/span>), then the number of length n<\/span> strings (shorted as n<\/span>-strings) is greater than n+8<\/span>.<\/p>\n

If you’ve established the simple<\/span> lemma, the solution is instant. Otherwise very impossible<\/span>.<\/p>\n

Lemma:<\/b> The number C_n<\/span> of n<\/span>-strings is strictly greater than C_{n-1}<\/span>, that of n-1<\/span>-strings.
\nActually, \u00a0we always have C_n \\ge C_{n-1}<\/span>: Every n-1<\/span>-string corresponds to an n<\/span>-string by continuing 1 more digit<\/span>. The map is clearly injective. If C_n=C_{n-1}<\/span>, it is bijective, meaning we have a way to continue uniquely<\/span>, which means rationality. Rigidly, at least one of the n-1<\/span>-strings occurs infinitely, but all digits after some n-1<\/span>-string is totally determined by it. So if an n-1<\/span>-string appears twice, it must appear every such distances, and so do the digits between.<\/p>\n

(Further discussion: For a rational number, split it into a finite part<\/span>, and a recurring part<\/span>. If the minimal length of recurring string is n<\/span>, then any m<\/span>-string starting in the recurring part has exactly n<\/span> variations, if m \\ge n<\/span>. Additional variations brought by the finite irregular part is finite (regardless of m<\/span>), as the starting point is finite. So C_n<\/span> in this case is not decreasing but bounded<\/span>. So it reaches some certain number and stays stable. In a purely recurring case, the number is exactly n<\/span> (meaning afore-defined).<\/p>\n

Now C_n \\ge C_{n-1}+1 \\ge ... \\ge 9+1<\/span>, as 10 > 9=1+8<\/span> holds, the problem is solved.<\/p>\n

This may not be really hard. But the most important thing is to see the principle behind it<\/span>.<\/p>\n

(I \u00a0WILL FURTHER COMPLETE THIS POST.)<\/p>\n

Saturday, August 10, 2013<\/p>\n

<\/p>","protected":false},"excerpt":{"rendered":"

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