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Question Number 194295 by Abdullahrussell last updated on 02/Jul/23

Answered by BaliramKumar last updated on 02/Jul/23

$$1!\times 2!\times 3!\times ..................\times 2023!$$$$1^{2023}\times 2^{2022}\times 3^{2021}\times ..................\times {2023}^1$$$$1^{2023}\times ...\times 5^{2019}\times ....\times {10}^{2014}\times ..................\times {2020}^4\times ...\times {2023}^1$$$$5^x\times y$$$$answer=x$$$$5^1\Rightarrow$$$$2019+2014+.......+4=\frac{n}{2}[a+a_n]=\frac{404}{2}[4+2019]$$$$202\times 2023=408646$$$$5^2\Rightarrow$$$${25}^{1999}+{50}^{1974}+..........+{2000}^{24}$$$$1999+1974+......+24=80920$$$$5^3\Rightarrow$$$${125}^{1899}+{250}^{1774}+.....+{2000}^{24}$$$$1899+1774+....+24=15384$$$$5^4\Rightarrow$$$${625}^{1399}+{1250}^{774}+{1875}^{149}$$$$1399+774+149=2322$$$$x=408646+80920+15384+2322$$$$\overline{)\begin{array}{c}x=507272\end{array}}$$

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