Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 138320 by aliibrahim1 last updated on 12/Apr/21

Answered by Rasheed.Sindhi last updated on 17/Apr/21

$${11}^n={(1+10)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right){\left(10\right)}^0+\left(\begin{array}{c}n\\ 1\end{array}\right){\left(10\right)}^1$$$$+\left(\begin{array}{c}n\\ 2\end{array}\right){\left(10\right)}^2+\underset{divisibleby1000}{\underbrace{\left(\begin{array}{c}n\\ 3\end{array}\right){\left(10\right)}^3....+\left(\begin{array}{c}n\\ n\end{array}\right){\left(10\right)}^n}}$$$$⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢$$$${\left(11\right)}^{9999}\equiv x\left(mod1000\right)$$$${(1+10)}^{9999}\equiv \left(\begin{array}{c}9999\\ 0\end{array}\right){\left(10\right)}^0$$$$+\left(\begin{array}{c}9999\\ 1\end{array}\right){\left(10\right)}^1$$$$+\left(\begin{array}{c}9999\\ 2\end{array}\right){\left(10\right)}^2\left(md1000\right)$$$$\equiv 1+99990+4998500100$$$$\equiv 4998600091\equiv 91$$

Commented by aliibrahim1 last updated on 27/Apr/21

$$thxman$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com