n^13 - n
2017/06/13 01:52
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標題:
發問:
對任何正整數n, p皆整除n^13 - n ,求p的最大值. 更新: 要用數字作答.
最佳解答:
1^13-1=02^13-2=8190=2×3^2×5×7×133^13-3=1594320=2^4×3×5×7×13×73So the common factor for n^13-n is at best 2×3×5×7×13Divisibility by 2: When n is odd, n^13 is odd, so n^13-n is evenwhen n is even n^13-n is even=>n^13-n is divisible by 2Divisibility by 3: (3n)^13-3n is divisible by 3(3n+1)^13-(3n+1)≡1-1 (mod 3)≡0 (mod 3)(3n-1)^13-(3n-1)≡-1+1 (mod 3)≡0 (mod 3)Divisibility by 5:(5n)^13-5n is divisible by 5(5n+1)^13-(5n+1)≡0 (mod 5)(5n-1)^13-(5n-1)≡0 (mod 5)(5n+2)^13-(5n+2)≡2^13-2 (mod 5)≡8190(mod 5)≡0(mod 5)(5n-2)^13-(5n-2)≡-2^13+2 (mod 5)≡0(mod 5)Divisibility by 7:(7n)^13-7n is divisible by 7(7n+1)^13-(7n+1)≡0 (mod 7)(7n-1)^13-(7n-1)≡0 (mod 7)(7n+2)^13-(7n+2)≡2^13-2 (mod 7)≡8190(mod 7)≡0(mod 7)(7n-2)^13-(7n-2)≡-2^13+2 (mod 7)≡0(mod 7)(7n+3)^13-(7n+3)≡3^13-3 (mod 7)≡1594320 (mod 7)≡0 (mod 7)(7n-3)^13-(7n-3)≡-3^13+3 (mod 7)≡0 (mod 7)Divisibility by 13:Let P(n) be the proposition that n^13-n is divisible by 13P(1) is obviosuly trueSuppose P(k) is true so k^13-k=13N(k+1)^13-(k+1)=k^13+13k^12+78k^11+286k^10+715k^9+1287k^8+1716k^7+1716k^6+1287k^5+715k^4+286k^3+78k^2+13k+1-k-1=k^13-k+13(k^12+6k^11+22k^10+55k^9+99k^8+132k^7+132k^6+99k^5+55k^4+22k^3+6k^2+k)=13(N+k^12+6k^11+22k^10+55k^9+99k^8+132k^7+132k^6+99k^5+55k^4+22k^3+6k^2+k)=>P(k+1) is true so n^13-n is divisible by 13So the largest value of p is 2×3×5×7×13=2730
其他解答:
費馬小定理解法 : n13 - n = n (n - 1) (n + 1) (n2 + 1) (n2 - n + 1) (n2 + n + 1) (n?- n2 + 1) 2013-01-02 02:06:42 補充： 包含因式 n (n - 1) = n2 - n ≡ 0 (mod 2) n (n - 1) (n + 1) = n3 - n ≡ 0 (mod 3) n (n - 1) (n + 1) (n2 + 1) = n^5 - n ≡ 0 (mod 5) n (n - 1) (n + 1) (n2 - n + 1) (n2 + n + 1) = n^7 - n ≡ 0 (mod 7) n13 - n ≡ 0 (mod 13) 故 2*3*5*7*13 = 2730 | n13 - n 。 最大性分析同自由自在~|||||用費馬小定理可以較快地得到結果
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n^13 - n發問:
對任何正整數n, p皆整除n^13 - n ,求p的最大值. 更新: 要用數字作答.
最佳解答:
1^13-1=02^13-2=8190=2×3^2×5×7×133^13-3=1594320=2^4×3×5×7×13×73So the common factor for n^13-n is at best 2×3×5×7×13Divisibility by 2: When n is odd, n^13 is odd, so n^13-n is evenwhen n is even n^13-n is even=>n^13-n is divisible by 2Divisibility by 3: (3n)^13-3n is divisible by 3(3n+1)^13-(3n+1)≡1-1 (mod 3)≡0 (mod 3)(3n-1)^13-(3n-1)≡-1+1 (mod 3)≡0 (mod 3)Divisibility by 5:(5n)^13-5n is divisible by 5(5n+1)^13-(5n+1)≡0 (mod 5)(5n-1)^13-(5n-1)≡0 (mod 5)(5n+2)^13-(5n+2)≡2^13-2 (mod 5)≡8190(mod 5)≡0(mod 5)(5n-2)^13-(5n-2)≡-2^13+2 (mod 5)≡0(mod 5)Divisibility by 7:(7n)^13-7n is divisible by 7(7n+1)^13-(7n+1)≡0 (mod 7)(7n-1)^13-(7n-1)≡0 (mod 7)(7n+2)^13-(7n+2)≡2^13-2 (mod 7)≡8190(mod 7)≡0(mod 7)(7n-2)^13-(7n-2)≡-2^13+2 (mod 7)≡0(mod 7)(7n+3)^13-(7n+3)≡3^13-3 (mod 7)≡1594320 (mod 7)≡0 (mod 7)(7n-3)^13-(7n-3)≡-3^13+3 (mod 7)≡0 (mod 7)Divisibility by 13:Let P(n) be the proposition that n^13-n is divisible by 13P(1) is obviosuly trueSuppose P(k) is true so k^13-k=13N(k+1)^13-(k+1)=k^13+13k^12+78k^11+286k^10+715k^9+1287k^8+1716k^7+1716k^6+1287k^5+715k^4+286k^3+78k^2+13k+1-k-1=k^13-k+13(k^12+6k^11+22k^10+55k^9+99k^8+132k^7+132k^6+99k^5+55k^4+22k^3+6k^2+k)=13(N+k^12+6k^11+22k^10+55k^9+99k^8+132k^7+132k^6+99k^5+55k^4+22k^3+6k^2+k)=>P(k+1) is true so n^13-n is divisible by 13So the largest value of p is 2×3×5×7×13=2730
其他解答:
費馬小定理解法 : n13 - n = n (n - 1) (n + 1) (n2 + 1) (n2 - n + 1) (n2 + n + 1) (n?- n2 + 1) 2013-01-02 02:06:42 補充： 包含因式 n (n - 1) = n2 - n ≡ 0 (mod 2) n (n - 1) (n + 1) = n3 - n ≡ 0 (mod 3) n (n - 1) (n + 1) (n2 + 1) = n^5 - n ≡ 0 (mod 5) n (n - 1) (n + 1) (n2 - n + 1) (n2 + n + 1) = n^7 - n ≡ 0 (mod 7) n13 - n ≡ 0 (mod 13) 故 2*3*5*7*13 = 2730 | n13 - n 。 最大性分析同自由自在~|||||用費馬小定理可以較快地得到結果
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