Results 1 to 10 of 10

Math Help - power problem

  1. #1
    Newbie
    Joined
    Feb 2008
    Posts
    13

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by spywx View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,676
    Thanks
    608
    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<br />
7^1 & \to & 7 \\<br />
7^2 & \to & 9 \\<br />
7^3 & \to & 3 \\<br />
7^4  & \to & 1\\<br />
\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<br />
11^1 & \to & 1 \\<br />
11^2 & \to & 1 \\<br />
11^3 & \to & 1 \\<br />
& \text{Hey!} \end{array}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2008
    Posts
    13

    Thumbs up

    Quote Originally Posted by Soroban View Post
    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<br />
7^1 & \to & 7 \\<br />
7^2 & \to & 9 \\<br />
7^3 & \to & 3 \\<br />
7^4  & \to & 1\\<br />
\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<br />
11^1 & \to & 1 \\<br />
11^2 & \to & 1 \\<br />
11^3 & \to & 1 \\<br />
& \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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2008
    Posts
    4
    Quote Originally Posted by spywx View Post
    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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2008
    From
    Brooklyn, NY USA
    Posts
    3
    Quote Originally Posted by ramelan View Post
    So you came on a forum asking for help on Prof. Boklan's question huh?

    ah the power of google XD
    x 2 LOL
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2008
    Posts
    13

    Cool

    Quote Originally Posted by ramelan View Post
    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 ?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Mar 2008
    From
    Brooklyn, NY USA
    Posts
    3
    I'm the short stocky/built white kid and he's the thin asian kid with glasses...Daniel and Brian.
    Last edited by SupraFast; March 3rd 2008 at 07:29 PM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Feb 2008
    Posts
    13
    Quote Originally Posted by SupraFast View Post
    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
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Mar 2008
    From
    Brooklyn, NY USA
    Posts
    3
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Power Problem
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: December 24th 2008, 12:01 AM
  2. Power Series Problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 7th 2008, 05:21 PM
  3. power sets problem
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: September 28th 2008, 06:12 AM
  4. problem of power series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 5th 2008, 05:29 AM
  5. Power problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 21st 2005, 02:02 AM

Search Tags


/mathhelpforum @mathhelpforum