1. ## power problem

hi this is my first post over here i wish i can get help with this

well i got this problem in my homework and i cant find a way to solve it
well the problem is
find the last digit of 7^1000000 and 11^1000000

so if plz if anyone has idea about how to solve it that would be great

thx

2. Originally Posted by spywx
hi this is my first post over here i wish i can get help with this

well i got this problem in my homework and i cant find a way to solve it
well the problem is
find the last digit of 7^1000000 and 11^1000000

so if plz if anyone has idea about how to solve it that would be great

thx
$\displaystyle 7^{100000}=49^{50000}=(40+9)^{50000}$

So the last digit of $\displaystyle 7^{100000}$ is the same as the last digit of $\displaystyle 9^{50000}$.

Now repeat the process:

$\displaystyle 9^{50000}=81^{25000}=(80+1)^{25000}$

So the last digit of $\displaystyle 9^{50000}$ is the same as the last digit of $\displaystyle 1^{25000}$ which is $\displaystyle 1$.

So the last digit of $\displaystyle 7^{100000}$ is $\displaystyle 1$.

RonL

3. Hello, spywx!

Welcome aboard!

Find the last digit of: $\displaystyle 7^{1,000,000}\text{ and }11^{1,000,000}$
Did you try anything?

I would try consecutive powers of 7 and see what happens . . .

$\displaystyle \begin{array}{ccc}\text{power} & & \text{ends in:} \\ \hline 7^1 & \to & 7 \\ 7^2 & \to & 9 \\ 7^3 & \to & 3 \\ 7^4 & \to & 1\\ \vdots & & \vdots \end{array}$

We see that $\displaystyle 7^4$ ends in 1.

We have: .$\displaystyle 7^{1,000,000} \;=\;(7^4)^{250,000} \:\to\:1^{250,000} \:\to\:1$

Try consecutive powers of 11 . . .

$\displaystyle \begin{array}{ccc}\text{power} & & \text{ends in:} \\ \hline 11^1 & \to & 1 \\ 11^2 & \to & 1 \\ 11^3 & \to & 1 \\ & \text{Hey!} \end{array}$

4. Originally Posted by Soroban
Hello, spywx!

Welcome aboard!

Did you try anything?

I would try consecutive powers of 7 and see what happens . . .

$\displaystyle \begin{array}{ccc}\text{power} & & \text{ends in:} \\ \hline 7^1 & \to & 7 \\ 7^2 & \to & 9 \\ 7^3 & \to & 3 \\ 7^4 & \to & 1\\ \vdots & & \vdots \end{array}$

We see that $\displaystyle 7^4$ ends in 1.

We have: .$\displaystyle 7^{1,000,000} \;=\;(7^4)^{250,000} \:\to\:1^{250,000} \:\to\:1$

Try consecutive powers of 11 . . .

$\displaystyle \begin{array}{ccc}\text{power} & & \text{ends in:} \\ \hline 11^1 & \to & 1 \\ 11^2 & \to & 1 \\ 11^3 & \to & 1 \\ & \text{Hey!} \end{array}$

no i did nt try this way b4 i was using totally different method, but your seems to be reasonable
so lets take another example i have 9^1000000 == (9^4)^250000

9^4 ==6561 last digit 1 then (9^4)^250000 last digit also must be 1

thx man

5. Originally Posted by spywx
no i did nt try this way b4 i was using totally different method, but your seems to be reasonable
so lets take another example i have 9^1000000 == (9^4)^250000

9^4 ==6561 last digit 1 then (9^4)^250000 last digit also must be 1

thx man
So you came on a forum asking for help on Prof. Boklan's question huh?

ah the power of google XD

6. Originally Posted by ramelan
So you came on a forum asking for help on Prof. Boklan's question huh?

ah the power of google XD
x 2 LOL

7. Originally Posted by ramelan
So you came on a forum asking for help on Prof. Boklan's question huh?

ah the power of google XD