Results 1 to 3 of 3

Math Help - Probability involving a number cubed and its last two digits

  1. #1
    Junior Member
    Joined
    Nov 2008
    Posts
    53

    Probability involving a number cubed and its last two digits

    So I'm not too sure about how to even approach this question:

    Suppose a number from 1 to 1000000 is selected at random. What is the probability that the last two digits of its cube are both 1?

    any help would be appreciated
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Shapeshift View Post
    So I'm not too sure about how to even approach this question:

    Suppose a number from 1 to 1000000 is selected at random. What is the probability that the last two digits of its cube are both 1?

    any help would be appreciated
    One simple approach boils the problem down to finding how many numbers from 1 to 1,000,000 have a cube that ends in "11". One way of doing this is to determine what the requirements are for the cube of an integer to end in "1" and then to determine the requirements for the cube of an integer to end in "11".

    Perfect cubes ending in "1": Looking at the cubes of the integers 1 through 12 it is seen that in order for the cube of an integer to end in "1", the integer must end in "1". So the numbers must be integers of the form 10x + 1.

    Perfect cubes ending in "11": From the above you know that the required numbers must have the form 10x + 1. (10x + 1)^3 = 1000x^3 + 300x^2 + 30x + 1 and in order for this expression to end in "11", we must have 30x + 1 ending in "11" (because the first two terms of the expression contribute nothing to the tens digit). For that to happen, the "30x" must have final digits "10" since the last term will add 1 to that. For that to happen, x must have final digit "7".

    Therefore you need to count all the numbers from 1 to 1,000,000 of the form 10x + 1 where the final digit of x is 7. The first of these numbers is given by x = 7 (and is 71) ....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Shapeshift View Post
    So I'm not too sure about how to even approach this question:

    Suppose a number from 1 to 1000000 is selected at random. What is the probability that the last two digits of its cube are both 1?

    any help would be appreciated
    Any number can be written in the form A=100u+x, where x is between 0 and 99

    So:

    A^3=(100u)^3+3(100u)^2x+3(100u)x^2+x^3

    and so for A^3 to end 11 requires that x^3 end in 11, but a quick check shows that there is only one such integer between 0 and 99 and that is 71.

    Can you take it from there?

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: May 3rd 2011, 03:22 PM
  2. [SOLVED] Combinatorics question involving summing digits of numbers
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: January 29th 2011, 06:21 AM
  3. Replies: 7
    Last Post: November 28th 2010, 09:22 PM
  4. Probability question involving number of permutations
    Posted in the Advanced Statistics Forum
    Replies: 10
    Last Post: January 15th 2010, 03:09 PM
  5. Replies: 3
    Last Post: April 7th 2008, 09:20 AM

Search Tags


/mathhelpforum @mathhelpforum