# Digis:Find it out

1) Find the last 3 digits of 9999...

## 3 Answers

1057
Ketan Chandak ·

using binomial....clearly....
9100â‰¡1(mod 1000)
now i think u can proceed further...

7
Sigma ·

no, Ketan,, i got stuck.. Please proceed further...

1708
man111 singh ·

\hspace{-16}\mathbb{F}$or the Calculation of Last$\bf{3}$Digit in$\bf{9^{9^{9^{9}}}}$, First We Will Calculate$\bf{9^9}$of\\\\ extreme Right Upper Part.\\\\ So$\bf{9^9=(10-1)^{9}=\binom{9}{0}.(10)^{10}-\binom{9}{1}.(10)^{9}+.....+\binom{9}{8}.(10)-\binom{9}{9}}$\\\\\\ So$\bf{9^9=\mathbb{M}(100)+90-1=89}$\\\\\\ Means Last$\bf{2}$Digit of$\bf{9^9}$is$\bf{=89}$, Where$\bf{\mathbb{M(\bf{100})}=M}$ultiple of$\bf{100}$\\\\\\ Now We Will Calculate$\bf{9^{89}}$\\\\\\ So$\bf{9^{89}=(10-1)^{89}=\binom{89}{0}.(10)^{10}-\binom{89}{1}.(10)^{9}+.......-\binom{89}{87}.(10)^2+\binom{89}{88}.(10)-\binom{89}{89}}$\\\\\\ So$\bf{9^{89}=\mathbb{M}(1000)-391600+890-1=\mathbb{M}(1000)+289}$\\\\\\ So Last$\bf{3}$Digit of$\bf{9^{89}}$is$\bf{289}$\\\\\\ Now Calculate$\bf{9^{289}}$\\\\\\ So$\bf{9^{289}=(10-1)^{289}=\binom{289}{0}.(10)^{10}-\binom{289}{1}.(10)^{9}+.......-\binom{289}{287}.(10)^2+\binom{289}{288}.(10)-\binom{289}{289}}$\\\\\\ \hspace{-16}$So $\bf{9^{289}=\mathbb{M}(1000)-4161600+2890-1=\mathbb{M(\bf{1000})+289}}$\\\\\\ So Last $\bf{3}$ Digit of $\bf{9^{9^{9^{9}}}}$is $\bf{\underline{\underline{289}}}$