Re: [問題] 98中山資工離散
2^n-1 = 2^0 + 2^1 +...
任何奇數m可以表示成 2^0 + 2^a + 2^b ...
則存在 2^n - 1 = ( 2^0 + .. ) + 2^c ( 2^0 + ... ) + ...
= m + 2^c * m + 2^2c * m + ...
德政 XD
※ 引述《a534055 (可樂)》之銘言：
: m是奇數 請用鴿籠原理證明
: 存在一個正整數n
: 使得m整除2^n-1
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