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

[Math100]

Homework Statement

Find the last two digits of the number $9^{9^{9}}$.
[Hint: $9^{9}\equiv 9\pmod {10}$; hence, $9^{9^{9}}=9^{9+10k}$; notice that $9^{9}\equiv 89\pmod {100}$.]

Relevant Equations

None.

Note that $9^{9}\equiv 9\pmod {10}$.
Thus
\begin{align*}
&9^{9^{9}}\equiv 9^{9+10k}\pmod {100}\\
&\equiv (9^{9}\cdot 9^{10k})\pmod {100}\\
&\equiv (9^{9}\cdot (-1)^{10k})\pmod {100}\\
&\equiv 9^{9}\cdot [(-1)^{10}]^{k}\pmod {100}\\
&\equiv (9^{9}\cdot 1)\pmod {100}\\
&\equiv 9^{9}\pmod {100}\\
&\equiv 89\pmod {100}.\\
\end{align*}
Therefore, the last two digits of the number $9^{9^{9}}$ are $89$.

 

[fresh_42]

Homework Statement:: Find the last two digits of the number $9^{9^{9}}$.
[Hint: $9^{9}\equiv 9\pmod {10}$; hence, $9^{9^{9}}=9^{9+10k}$; notice that $9^{9}\equiv 89\pmod {100}$.]
Relevant Equations:: None.

Note that $9^{9}\equiv 9\pmod {10}$.
Thus
\begin{align*}
&9^{9^{9}}\equiv 9^{9+10k}\pmod {100}\\
&\equiv (9^{9}\cdot 9^{10k})\pmod {100}\\
&\equiv (9^{9}\cdot (-1)^{10k})\pmod {100}\\
&\equiv 9^{9}\cdot [(-1)^{10}]^{k}\pmod {100}\\
&\equiv (9^{9}\cdot 1)\pmod {100}\\
&\equiv 9^{9}\pmod {100}\\
&\equiv 89\pmod {100}.\\
\end{align*}
Therefore, the last two digits of the number $9^{9^{9}}$ are $89$.

I do not understand the third step, where you go from $9^{10k}$ to $(-1)^{10k}.$ Why should this be true modulo $100$? It is true modulo $10$, not $100.$

 

[Math100]

I do not understand the third step, where you go from $9^{10k}$ to $(-1)^{10k}.$ Why should this be true modulo $100$? It is true modulo $10$, not $100.$

I think I am wrong, then. Because I thought it is true modulo $100$.

 

[fresh_42]

I think I am wrong, then. Because I thought it is true modulo $100$.

Try $(9^{10})^k=(9^9\cdot 9)^k$ modulo $100$.

 

[Math100]

Try $(9^{10})^k=(9^9\cdot 9)^k$ modulo $100$.

$(9^{10})^{k}\equiv 1^{k}\pmod {100}$

 

[Math100]

Because $9^{10}\equiv 1\pmod {100}$.

 

[Math100]

Observe that $9^{9}\equiv 9\pmod {10}$ and $9^{10}\equiv 1\pmod {100}$.
Thus
\begin{align*}
&9^{9^{9}}\equiv (9^{9+10k})\pmod {100}\\
&\equiv (9^{9}\cdot 9^{10k})\pmod {100}\\
&\equiv [9^{9}\cdot (1)^{k}]\pmod {100}\\
&\equiv (9^{9}\cdot 1)\pmod {100}\\
&\equiv 9^{9}\pmod {100}\\
&\equiv 89\pmod {100}.\\
\end{align*}
Therefore, the last two digits of the number $9^{9^{9}}$ are $89$.

 

[fresh_42]

Observe that $9^{9}\equiv 9\pmod {10}$ and $9^{10}\equiv 1\pmod {100}$.
Thus
\begin{align*}
&9^{9^{9}}\equiv (9^{9+10k})\pmod {100}\\
&\equiv (9^{9}\cdot 9^{10k})\pmod {100}\\
&\equiv [9^{9}\cdot (1)^{k}]\pmod {100}\\
&\equiv (9^{9}\cdot 1)\pmod {100}\\
&\equiv 9^{9}\pmod {100}\\
&\equiv 89\pmod {100}.\\
\end{align*}
Therefore, the last two digits of the number $9^{9^{9}}$ are $89$.

Yes, but how did you find $9^{10}\equiv 1 \pmod {10}$?

Can you explain that and why $9^9\equiv 9 \pmod {10}$ without calculating the products?

 

[Math100]

Yes, but how did you find $9^{10}\equiv 1 \pmod {10}$?

Can you explain that and why $9^9\equiv 9 \pmod {10}$ without calculating the products?

For $9^{9}\equiv 9\pmod {10}$, I got this from the hint in this question/problem. As for $9^{10}\equiv 1\pmod {100}$, I found that by,

$9^{10}\equiv (9^{9}\cdot 9)\pmod {100}\equiv (89\cdot 9)\pmod {100}\equiv 801\pmod {100}\equiv 1\pmod {100}$.
Thus $9^{10}\equiv 1\pmod {100}$.

 

[fresh_42]

For $9^{9}\equiv 9\pmod {10}$, I got this from the hint in this question/problem. As for $9^{10}\equiv 1\pmod {10}$, I found that by,

$9^{10}\equiv (9^{9}\cdot 9)\pmod {100}\equiv (89\cdot 9)\pmod {100}\equiv 801\pmod {100}\equiv 1\pmod {100}$.
Thus $9^{10}\equiv 1\pmod {100}$.

Very well, that was the idea. Yes, $9^9\equiv 89 \pmod {100}$ needs probably a computation ($9^3=729$ and $29^3=24,389$) and $9^9\equiv 1 \pmod {10}$ is a consequence thereof. The latter can also be seen by:
\begin{align*}
9^1&\equiv 9 \pmod {10}\\
9^2&\equiv 1 \pmod {10}\\
9^3&\equiv 9 \pmod {10}\\
9^4&\equiv 1 \pmod {10}\\
9^5&\equiv 9 \pmod {10}\\
&\ldots
\end{align*}
The digits $1$ and $9$ alternate.