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Question Number 108781 by 150505R last updated on 19/Aug/20

Answered by 1549442205PVT last updated on 19/Aug/20

$$\mathrm{P}=(1!\times 1)+(2!\times 2)+(3!\times 3)+...+(286!\times 286)$$$$=[1!\times (2-1)]+[2!\times (3-1)]+[3!\times (4-1)]+(4!\times (5-1)]+$$$$...+(286!\times (287-1)]$$$$=-1!+2!-2!+3!-3!+4!-4!+5!-...$$$$-286!+287!=287!$$$$\mathrm{Hence}\mathrm{P}=287!\mathrm{It}\mathrm{follows}\mathrm{that}$$$$\frac{\mathrm{P}}{2^4\times 5^3}=\frac{287!}{2^4\times 5^3}=\frac{287!}{2\times {10}^3}=\frac{287!}{2000}$$$$\mathrm{Since}287!=1\times 2\times 3\times 4\times 5\times ....\times 287$$$$\mathrm{which}\mathrm{contain}\mathrm{more}3\mathrm{factors}5\mathrm{and}$$$$\mathrm{more}4\mathrm{factors}2,\mathrm{so}\mathrm{P}=0\mathrm{mod}\left(2^4\times 5^3\right)$$$$\mathrm{Thus},\mathrm{P}\mathrm{divide}\mathrm{by}\left(2^4\times 5^3\right)\mathrm{gives}\mathrm{the}$$$$\mathrm{remainder}\mathrm{equal}\mathrm{to}\mathrm{zero}$$$$$$

Commented by 150505R last updated on 19/Aug/20

$$pis287!-1$$

Commented by 1549442205PVT last updated on 19/Aug/20

$$\mathrm{Thank}\mathrm{you}.\mathrm{I}\mathrm{missed}(-1).\mathrm{We}\mathrm{have}$$$$287!-1=-1\mathrm{mod}\left(2^4\times 5^3\right)$$

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