Question Number 108781 by 150505R last updated on 19/Aug/20
Answered by 1549442205PVT last updated on 19/Aug/20
![P=(1!×1)+(2!×2)+(3!×3)+...+(286!×286) =[1!×(2−1)]+[2!×(3−1)]+[3!×(4−1)]+(4!×(5−1)]+ ...+(286!×(287−1)] =−1!+2!−2!+3!−3!+4!−4!+5!−... −286!+287!=287! Hence P=287! It follows that (P/(2^4 ×5^3 ))=((287!)/(2^4 ×5^3 ))=((287!)/(2×10^3 ))=((287!)/(2000)) Since 287!=1×2×3×4×5×....×287 which contain more 3 factors 5 and more 4 factors 2 ,so P=0 mod(2^4 ×5^3 ) Thus ,P divide by (2^4 ×5^3 ) gives the remainder equal to zero](https://www.tinkutara.com/question/Q108800.png)
$$\mathrm{P}=\left(\mathrm{1}!×\mathrm{1}\right)+\left(\mathrm{2}!×\mathrm{2}\right)+\left(\mathrm{3}!×\mathrm{3}\right)+…+\left(\mathrm{286}!×\mathrm{286}\right) \\ $$$$=\left[\mathrm{1}!×\left(\mathrm{2}−\mathrm{1}\right)\right]+\left[\mathrm{2}!×\left(\mathrm{3}−\mathrm{1}\right)\right]+\left[\mathrm{3}!×\left(\mathrm{4}−\mathrm{1}\right)\right]+\left(\mathrm{4}!×\left(\mathrm{5}−\mathrm{1}\right)\right]+ \\ $$$$…+\left(\mathrm{286}!×\left(\mathrm{287}−\mathrm{1}\right)\right] \\ $$$$=−\mathrm{1}!+\mathrm{2}!−\mathrm{2}!+\mathrm{3}!−\mathrm{3}!+\mathrm{4}!−\mathrm{4}!+\mathrm{5}!−… \\ $$$$−\mathrm{286}!+\mathrm{287}!=\mathrm{287}! \\ $$$$\mathrm{Hence}\:\mathrm{P}=\mathrm{287}!\:\:\:\mathrm{It}\:\mathrm{follows}\:\mathrm{that} \\ $$$$\frac{\mathrm{P}}{\mathrm{2}^{\mathrm{4}} ×\mathrm{5}^{\mathrm{3}} }=\frac{\mathrm{287}!}{\mathrm{2}^{\mathrm{4}} ×\mathrm{5}^{\mathrm{3}} }=\frac{\mathrm{287}!}{\mathrm{2}×\mathrm{10}^{\mathrm{3}} }=\frac{\mathrm{287}!}{\mathrm{2000}} \\ $$$$\mathrm{Since}\:\mathrm{287}!=\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}×\mathrm{5}×….×\mathrm{287} \\ $$$$\:\mathrm{which}\:\mathrm{contain}\:\mathrm{more}\:\mathrm{3}\:\mathrm{factors}\:\mathrm{5}\:\mathrm{and}\: \\ $$$$\mathrm{more}\:\mathrm{4}\:\mathrm{factors}\:\mathrm{2}\:,\mathrm{so}\:\mathrm{P}=\mathrm{0}\:\mathrm{mod}\left(\mathrm{2}^{\mathrm{4}} ×\mathrm{5}^{\mathrm{3}} \right) \\ $$$$\mathrm{Thus}\:,\mathrm{P}\:\mathrm{divide}\:\mathrm{by}\:\left(\mathrm{2}^{\mathrm{4}} ×\mathrm{5}^{\mathrm{3}} \right)\:\mathrm{gives}\:\mathrm{the} \\ $$$$\mathrm{remainder}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{zero} \\ $$$$ \\ $$
Commented by 150505R last updated on 19/Aug/20
$${p}\:{is}\:\mathrm{287}!−\mathrm{1} \\ $$
Commented by 1549442205PVT last updated on 19/Aug/20
$$\mathrm{Thank}\:\mathrm{you}.\mathrm{I}\:\mathrm{missed}\:\left(−\mathrm{1}\right).\mathrm{We}\:\mathrm{have} \\ $$$$\mathrm{287}!−\mathrm{1}=−\mathrm{1mod}\left(\mathrm{2}^{\mathrm{4}} ×\mathrm{5}^{\mathrm{3}} \right) \\ $$