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Question Number 119097 by ZiYangLee last updated on 22/Oct/20

$$\mathrm{Find}\mathrm{the}\mathrm{remainder}\mathrm{when}$$$$(1!\times 1)+(2!\times 2)+(3!\times 3)+...+(100!\times 100)$$$$\mathrm{is}\mathrm{divided}\mathrm{by}101.$$

Answered by mindispower last updated on 22/Oct/20

$$n!.n=(n!)(n+1-1)=(\mathit{n}+1)!-n!$$$$\mathrm{\Sigma }k!.k=\underset{k=1}{\sum ^{100}}((k+1)!-k!)=101!-1!$$$$=101!-1=-1\left(101\right)=100\left(101\right)$$$$remainder=100$$$$$$

Answered by 1549442205PVT last updated on 22/Oct/20

$$\mathrm{Put}\mathrm{S}=1.1!+2.2!+3.3!+...+100.100!$$$$=1.1!+(3-1)2!+(4-1)3!+...+(101-100)100!$$$$=1.1!+(3!-2!)+(4!-3!)+(5!-4!)+...$$$$+(101!-100!)=1!-2!+101!=101!-1$$$$\mathrm{Thus},\mathrm{the}\mathrm{remainer}\mathrm{when}\mathrm{S}\mathrm{is}\mathrm{divided}$$$$\mathrm{by}101\mathrm{equal}\mathrm{to}(-1)$$

Commented by PRITHWISH SEN 2 last updated on 22/Oct/20

$$-1{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{be}\mathrm{a}\mathrm{remainder}\mathrm{it}\mathrm{will}\mathrm{be}\mathrm{written}\mathrm{as}$$$$100-101=100\mathrm{as}\mathrm{remainder}$$

Commented by 1549442205PVT last updated on 25/Oct/20

$$\mathrm{In}\mathrm{integer}\mathrm{numbers}\mathrm{it}\mathrm{is}\mathrm{always}\mathrm{possible}$$$$\mathrm{n}=3\mathrm{k}\pm 1...,\mathrm{Sir}!$$

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