求1*3*5*7*...*1997的末三位数字中的问题 求M=1*3*5*...*1997模1000的余数又因为125|M 所以M=125m 所以考虑模8的余数因为(2n-3)(2n-1)(2n+1)(2n+3)≡1(mod8)因为M是连续999个奇数乘积切999=4*249+3得M≡1*3*5≡7(mod8) (
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![求1*3*5*7*...*1997的末三位数字中的问题 求M=1*3*5*...*1997模1000的余数又因为125|M 所以M=125m 所以考虑模8的余数因为(2n-3)(2n-1)(2n+1)(2n+3)≡1(mod8)因为M是连续999个奇数乘积切999=4*249+3得M≡1*3*5≡7(mod8) (](/uploads/image/z/11456285-5-5.jpg?t=%E6%B1%821%2A3%2A5%2A7%2A...%2A1997%E7%9A%84%E6%9C%AB%E4%B8%89%E4%BD%8D%E6%95%B0%E5%AD%97%E4%B8%AD%E7%9A%84%E9%97%AE%E9%A2%98+%E6%B1%82M%3D1%2A3%2A5%2A...%2A1997%E6%A8%A11000%E7%9A%84%E4%BD%99%E6%95%B0%E5%8F%88%E5%9B%A0%E4%B8%BA125%7CM+%E6%89%80%E4%BB%A5M%3D125m+%E6%89%80%E4%BB%A5%E8%80%83%E8%99%91%E6%A8%A18%E7%9A%84%E4%BD%99%E6%95%B0%E5%9B%A0%E4%B8%BA%282n-3%29%282n-1%29%282n%2B1%29%282n%2B3%29%E2%89%A11%28mod8%29%E5%9B%A0%E4%B8%BAM%E6%98%AF%E8%BF%9E%E7%BB%AD999%E4%B8%AA%E5%A5%87%E6%95%B0%E4%B9%98%E7%A7%AF%E5%88%87999%3D4%2A249%2B3%E5%BE%97M%E2%89%A11%2A3%2A5%E2%89%A17%28mod8%29+%28)
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