Maths(HKMO)
2010-01-02 8:12 pm
If R is the remainder of 1^6+2^6+3^6+4^6+5^6+6^6 divided by 7, find the value of R.
回答 (2)
2010-01-02 8:29 pm
✔ 最佳答案
1^6+2^6+3^6+4^6+5^6+6^6 (mod 7)=1+64+2^3+2^3+(-2)^6+(-1)^6 (mod 7)
=1+64+8+8+64+1 (mod 7)
=146 (mod 7)
=6
R=6
2010-01-04 12:21 am
x^6+y^6 = (x+y)(x^5-x^4y+x^3 y^2-x^2 y^3+xy^4-y^5)+2y^6
6^6+1^6 = 7Q1+2
5^6+2^6 = 7Q2+2x2^6
4^6+3^6 = 7Q3+2x3^6
2+2x2^6+2x3^6
=2(1+64+729)
=1588
=7x226+6
The value of R is 6.
6^6+1^6 = 7Q1+2
5^6+2^6 = 7Q2+2x2^6
4^6+3^6 = 7Q3+2x3^6
2+2x2^6+2x3^6
=2(1+64+729)
=1588
=7x226+6
The value of R is 6.
收錄日期: 2021-04-23 18:26:59
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100102000051KK00521