power problem

**spywx** · February 28th 2008, 09:58 PM

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

**CaptainBlack** · February 28th 2008, 10:35 PM

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

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

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

Now repeat the process:

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

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

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

RonL

**Soroban** · February 29th 2008, 06:31 AM

Hello, spywx!

Welcome aboard!

Find the last digit of: $7^{1,000,000}\text{ and }11^{1,000,000}$

Did you try anything?

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

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

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

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

Try consecutive powers of 11 . . .

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

**spywx** · February 29th 2008, 07:42 AM

Originally Posted by Soroban

Hello, spywx!

Welcome aboard!

Did you try anything?

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

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

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

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

Try consecutive powers of 11 . . .

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

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

**ramelan** · March 2nd 2008, 07:15 PM

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

**SupraFast** · March 2nd 2008, 07:16 PM

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

**spywx** · March 2nd 2008, 11:34 PM

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

lol yeah the power of google u know its addictive, i treat google like my big bro i ask it 4 help and always find answer
i wonder who r u ?

**SupraFast** · March 3rd 2008, 05:37 PM

I'm the short stocky/built white kid and he's the thin asian kid with glasses...Daniel and Brian.

**spywx** · March 3rd 2008, 07:39 PM

Originally Posted by SupraFast

I'm the short stocky/built white kid and he's the thin asian kid with glasses...Daniel and Brian.

man fr a sec i thought he's the prof xD , nice 2 see u here and in class 2
btw i got a warning bcoz of ur friend

**SupraFast** · March 6th 2008, 10:24 PM

So who are you then? A warning from who? There shouldn't be anything wrong with asking for help on something you don't understand.

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