试解 |
送交者: zhf 2020月07月04日00:38:23 于 [灵机一动] 发送悄悄话 |
回 答: 趣味的数学-383 由 gugeren 于 2020-07-02 22:53:01 |
解:
3: 求2^m的3余数 2^0, r = 1 2^1, r = 2 2^2, r = 1 循环。所以 2^(2n), r = 1。 2^1000=2^(2(500)), r = 1 被3除之后的余数=1 5: 求2^m的5余数 2^0, r = 1 2^1, r = 2 2^2, r = 4 2^3, r = 3 2^4, r = 1 循环。所以 2^(4n), r = 1。 2^1000=2^(4(250)), r = 1 被5除之后的余数=1 7: 求2^m的7余数 2^0, r = 1 2^1, r = 2 2^2, r = 4 2^3, r = 1 循环。所以 2^(3n), r = 1。 2^999=2^(3(333)), r = 1 2^1000, r = 2 被7除之后的余数=2 11: 求2^m的11余数 2^0, r = 1 2^1, r = 2 2^2, r = 4 2^3, r = 8 ... 2^10, r = 1 循环。所以 2^(10n), r = 1 2^1000 = 2^(10(100)), r = 1 被11除之后的余数=1 13: 求2^m的13余数 2^(12n), r = 1 2^(12(83))=2^996, r = 1 2^997, r = 2 2^998, r = 4 2^999, r = 8 2^1000, r = 3 被13除之后的余数=3 17: 求2^m的17余数 2^(8n), r = 1 2^1000 = 2^(8(125)), r = 1 被17除之后的余数=1 19: 求2^m的19余数 2^(18n), r = 1 2^(18(55))= 2^990, r = 1 2^1000, r = 17 被19除之后的余数=17 |
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