求1^3+11^3+21^3+…+20091^3+20101^3除以2009時的餘數。
答案:1332
一條求餘數的奧數問題…
2010-07-26 6:59 am
回答 (4)
2010-07-26 11:09 am
✔ 最佳答案
求1^3+11^3+21^3+…+20091^3+20101^3除以2009時的餘數。Sol1^3+11^3+21^3+…+20091^3+20101^3=(1^3+11^3)+21^3+…+20091^3+20101^3=(1^3+11^3)+Σ(k=1 to 2009)_(10k+11)^3=1332+Σ(k=1 to 2009)_(ak^3+bk^2+ck+d)=1332+Σ(k=1 to 2009)_ak^3+Σ(k=1 to 2009)_bk^2+Σ(k=1to 2009)_ck+Σ(k=1 to 2009)_d餘數=1332
2010-07-26 6:10 pm
I think it is non-obvious why we have
Σ(k=1 to 2009)_ak^3 = 0 (mod 2009)
can you explain?
Σ(k=1 to 2009)_ak^3 = 0 (mod 2009)
can you explain?
2010-07-26 7:13 am
To B Hong:
即係1..+11..+21..+31..+41如此類推
即係1..+11..+21..+31..+41如此類推
2010-07-26 7:09 am
1^3+11^3+21^3+…+20091^3+20101^3 個T(n)系咩= =??
搞吾清楚個"+...+"做吾到
搞吾清楚個"+...+"做吾到
收錄日期: 2021-04-15 22:43:40
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