5. Does there exist a positive integer n such that n has exactly 2000 prime divisors and n divides 2"+1?(RUS)5.确定是否存在满足下列条件的正整数n:n恰好能够被2000个互不相同的质数整除,且 2^n+1 能够被n整除.(俄罗斯) 相关知识点:
Numbers n such that n divides 2^n + 1.(Formerly M2806)Select[Range[10^5], Divisible[2^# + 1, #] &] (* 注：效率不如 PowerMod. *)} 除了形如3^k, 还有其他的解：形如 a*3^k的，除了a=1， 还有a=19*9=171，a=3249=19^2*9，a=13203=163*9*9，... 待研究。Select[1 + ...
1） B is divisible by A : B/A with remainder 02) A divided by B : A/B3) A divides B : B/A4) A is divisor of B : B/Aby后面那个是分母 个人看法
,a(k≥2)be distinct integers in the set {1,…,n} such that n divides a;(a+1-1)for i-1,…,k-1. Prove that n does not divide ap(a1-1). (AUS) 1.设n是一个正整数,a1,a2,…,ak(k≥2)是集合{1,…,n}中的互不相同的整数,使得对 于 i=1,…,k-1,都有n整除a;(ai+1-1)...
设a1a2……an是任意正整数,证明:存在i在k(i>=0,k>=1)使得ai+1 + ai+2 +……+ai+k能被n整除 试证明:4^k-1一定能被3整除,其中k为正整数. 证明所有k,n属于整数,(k-n)能被(k-1)整除当且仅当(k-n)能被(n-1)整除.英文原题:For all k,n in Z,(k-n) divides (k-1) if only if ...
1. Let n be a positive integer and let a,., a_k(k≥2) be distinct integers in the set (1,⋯,n)such that n divides a, (a,+1-1) for i =1,…, k-1. Prove that n does not divide a_k(a_1-1)(AUS)1.设n是一个正整数,a1, a_2 ,…, a_k(k≥2) 是集合 (1,...
Adds 1 CR 1- ( nu1 -- nu2 ) Subtracts 1 CR 2+ ( nu1 -- nu2 ) Adds 2 CR 2- ( nu1 -- nu2 ) Subtracts 2 59 2* ( x1 -- x2 ) Multiplies by 2 57 2/ ( x1 -- x2 ) Divides by 2 27 lshift ( x1 u -- x2 ) Left shifts x1 ...
Remark The multiplication by the factor 2(n+2)2(n+2) in the 2nd last step is not ad-hoc. Rather, it corresponds to taking the "norm" of a quadratic number, since here we have that n2≡−1/2n2≡−1/2 so, ring-theoretically, nn behaves like −1/2−−−−√=−2...
故答案为B。 (2)C.考查动词及语境理解。 A.knew知道; B .learned学习 ; C .noticed注意到; D. acknow ledged承认。 根据下文I have a line of apple tre es running alongside the fence可知,作者觉得 他邻居应该注意到了沿着篱笆有一排苹果树。 故答 案为C。 (3)C.考查动词及语境理解。 A ....
Proof.(a^2+3b^2)(c^2+3d^2)=(ac+3bd)^2 + 3(bc-ad)^2 Additonal: Fermat's Last Theorem: n = 3 Lemma: if p,q are coprime, p,q opposite parity, then gcd(2p,p2 + 3q2) = 1 or 3 (1) Assume that there is a prime f which divides both 2p and p2 + 3q2. ...