The Base of 9

$\begin{aligned} N & = 99999^2 \\ & = (10^5 -1)^2 \\ & = 10^{10}-2\times 10^5 + 1 \\ & = 10,000,000,000 - 200,000 + 1 \\ & = \boxed{9,999,800,001} \end{aligned}$

There is a algorithm.

$\large9^{2}$ = $\large81$

$\large99^{2}$ = $\large9801$

$\large999^{2}$ = $\large998001$

$\large9999^{2}$ = $\large99980001$

$\large99999^{2}$ = $\large9999800001$

From the above algorithm we have got that 9 in exponent more than 1 time that repeat give a (9) at first and a zero between (81).

Example=

$\large99^{2}$ = $\large9801$ got a (9)(0).

hence the aswer is $\large9999800001$ .

$99999^2$

$=99999*99999$

$=99999*(100000-1)$

$=9999900000-99999$

$=\boxed{9999800001}$

By Ekanyunena Purvena Sutra of Vedic Mathematics

• 1st five digits of answer will be 99999-1=99998
• remaining 5 digits will be 100000-99999=00001

Ans: 9999800001