不要被那个深奥的梅森素数吓着了,其实和它没有什么太大关系的。 |
Consider 2^n where n=3,4,5... The last digits of the sequence are 8,6,2,4,8,6,2,4,... The 4 numbers repeat. So 2^p-1 cannot have 9 as the last digit. In order to get 3, 2^p must be 4. But when 2^n has 4 as the last digit, n is always even. |
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