Find the last two digits of the number $9^{9^9}$ [duplicate]

Find the last two digits of the number $9^{9^9}$ .

[Hint: $9^9 \equiv 9 \pmod {10} $; hence, $9^{9^9}$ = $9^9+10k$ ;now use the fact that $9^9 \equiv 89 \pmod {100}$]

Solution 1:

Our observations start like this;

$9=-1 ($mod $10) \Rightarrow 9^9=(-1)^9 ($mod $10)$ or $9^9=-1 $(mod $10)$ We have $-1=9 ($mod $10)$, thus $9^9=9 ($mod $10)$ ...i

This statement directly implies that upon dividing $9^9$ by $10$ we get a remainder $9$.Thus $9^9=9+10k$

Also,$9^9=89 ($mod $100) \Rightarrow 9*9^9=9*89 ($mod $100)$ or $9^{10}=801($mod $100)$. Also, $801=1 ($mod $100)$ thus making $9^{10} = 1 \pmod{100}$ ....ii

$(9^{10})^k=1^k ($mod $100)$ from ii $9^9*(9^{10})^k=9^9 ($mod $100)$ or, $9^{(9+10k)}=9^9 ($mod $100)$. Since,$9^9=89 ($mod $100)$, we have $9^{(9+10k)}=89 ($mod $100)$ This statement implies that upon division of $9^{9^9}$ by $100$ we get a remainder of $89$. Thus last two digits are $89$.

Solution 2:

$$9^{9^9}=(10-1)^{9^9}\equiv(-1)^{9^9}+9^9\cdot10^1\cdot(-1)^{9^9-1}\equiv-1+10\cdot9^9\pmod{100}$$

Now, $\displaystyle9^9=(10-1)^9\equiv-1\pmod{10}\implies10\cdot9^9\equiv-10\pmod{10\cdot10}$

$$\implies9^{9^9}\equiv-1-10\pmod{100}\equiv100-11$$