Okay so if you are given $\displaystyle 7^{999}$ how would you figure out the last two digits of the munber using mod.

- Aug 2nd 2009, 10:04 AMdiddledabble[SOLVED] Finding the last two digits of a number raised to a large exponent
Okay so if you are given $\displaystyle 7^{999}$ how would you figure out the last two digits of the munber using mod.

- Aug 2nd 2009, 10:44 AMrunning-gag
- Aug 2nd 2009, 10:48 AMdiddledabble
But how do you work through it. I did it for $\displaystyle 7^9999$ and got 143 but I am not confident on my steps. I don't want anyone to solve my homework. That is why I altered my question to $\displaystyle 7^999$. I just need a good detailed step by step explanation. Thanks

- Aug 2nd 2009, 12:22 PMo_O
Use Euler's theorem: $\displaystyle (a,m) = 1 \ \Rightarrow a^{\phi(m)} \equiv 1 \ (\text{mod } m)$ where $\displaystyle m \geq 1$

So we know: $\displaystyle \phi (100) = \phi \left(2^2 \cdot 5^2\right) = \cdots = 40$

Thus: $\displaystyle 7^{40} \equiv 1 \ (\text{mod } 100)$

Now use the property: $\displaystyle a \equiv b \ (\text{mod } m) \ \Rightarrow \ a^n \equiv b^n \ (\text{mod } m)$ - Aug 2nd 2009, 12:52 PMSoroban
Hello, diddledabble!

Quote:

Find the last two digits of: .$\displaystyle 7^{999}$

. . $\displaystyle \begin{array}{c|c}

\text{Power} & \text{Ends in} \\ \hline

7^1 & 07 \\ 7^2 & 49 \\ 7^3 & 43 \\ 7^4 & 01 \\ 7^5 & 07 \\ 7^6 & 49 \\ \vdots & \vdots \end{array}$

The endings form a four-step cycle: .$\displaystyle \{07, 49, 43, 01\}$

Since $\displaystyle 999 \equiv 3 \text{(mod 4)},\;7^{999}$ has the third ending: .$\displaystyle 43$

- Aug 4th 2009, 01:28 PMdiddledabble
Soroban,

Your answer was sooo easy to understand. Just what I was looking for.