摘要
借助中国剩余定理探讨Fermat数的尾数,证明了当非负整数n≥3时,Fermat数Fn=22n+1≡17,257,537,297,617,457,937,97,217,657,337,897,817,857,737,697,417,57,137,497(mod 1000).
By using Chinese remainder theorem,the mantissa of Fermat numbers is studied.And a conclusion that if n≥3,then Fermat number Fn=22n+1≡17,257,537,297,617,457,937,97,217,657,337,897,817,857,737,697,417,57,137,497(mod 1000) is given,where n is positive integer.
出处
《山东理工大学学报（自然科学版）》
CAS
2011年第5期47-48,共2页
Journal of Shandong University of Technology:Natural Science Edition
