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Is There an Easier Way to Solve This?

  1. Jul 21, 2004 #1

    loseyourname

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    Suppose a number between 1 and 100 is chosen at random. What is the probability that the last two digits of its cube are both 1?

    [tex]n = (1, 2, . . ., 1000)[/tex]

    [tex]n = x + y, \ x = (100, 200, . . ., 900), \ y = (1, 2, . . ., 100)[/tex]

    [tex]n^3 = (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3[/tex] =

    [tex](x^3 + 3x^2y + 3xy^2) + y^3[/tex] =

    [tex][x(x^2 + 3xy + 3y^2)] + y^3[/tex]

    Given that [itex]x = (100, 200, . . ., 900)[/itex] the last two digits of [itex]xz[/itex] will be 00, so long as [itex]z[/itex] is a whole number.

    Let [tex]z = x^2 + 3xy + 3y^2[/tex]

    Because [itex]x[/itex] and [itex]y[/itex] are both whole numbers, [itex]z[/itex] will also be a whole number. Therefore, the last two digits of [itex]x(x^2 + 3xy + 3y^2)[/itex] are 00. (1)

    Because [itex]n^3 = (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3[/itex], the last two digits of [itex]n^3[/itex] depend solely on [itex]y^3[/itex] by (1). Therefore, I only need to calculate the probability that the last two digits of [itex]y^3[/itex] are 11.

    Let [itex]A = [/itex] the last two digits of [itex]y^3[/itex] are 11.
    [itex]N[/itex] = all possible values of [itex]y[/itex] = 100.
    [itex]N(A)[/itex] = all values of [itex]y[/itex] such that [itex]y^3 = 100a + 11[/itex] (because the last two digits of [itex]n^3[/itex] independent of [itex]y^3[/itex] are 00).

    [tex]y^3 = 100a + 11[/tex]

    [tex]y = \sqrt[3]{100a + 11}[/tex]

    Clearly there is only one solution to this equation, so [itex]N(A)[/itex] = 1, so

    [tex]P(A) = \frac{N(A)}{N}[/tex] = 0.01
     
    Last edited: Jul 21, 2004
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  3. Jul 21, 2004 #2

    loseyourname

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    You know, looking at this now, the solution doesn't look all that complicated, but it took me almost an hour to figure it out last night.
     
  4. Jul 22, 2004 #3
    A lot of what you've done can be simplified if you use modulo arithmetic. You want to find the number of solutions to [tex]x^3 \equiv 11 \pmod{100}[/tex] where [tex]1 \leq x \leq 100[/tex]. Suppose [tex]x[/tex] is such a number that fulfils both equations. Then (by the very definition of modulo) there is an integer [tex]k[/tex] such that [tex]x^3 = 100k + 11[/tex], and apparently, one can immediatly see that this equation only has one solution (I don't understand how, but I'll take your word for it ;)).
     
  5. Jul 22, 2004 #4

    loseyourname

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    I should have specified that y must be a whole number between 1 and 100. Only one whole number between 1 and 100 can be represented that way (71). "Clearly" is a little misleading.

    By the way, I barely began to learn about congruence arithmetic. I never would have thought of that.
     
  6. Jul 22, 2004 #5

    NateTG

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    Well, it's a cubic in a field, so it has at most three, which is certainly close enough.

    Alternatively:
    You can also start by just looking at the last digit (i.e. residues mod 10):
    [tex]0^3=0[/tex]
    [tex]1^3=1[/tex]
    [tex]2^3=8[/tex]
    [tex]3^3=27[/tex]
    [tex]4^3=64[/tex]
    [tex]5^3=125[/tex]
    [tex]6^3=216[/tex]
    [tex]7^3=343[/tex]
    [tex]8^3=512[/tex]
    [tex]9^3=729[/tex]
    Clearly for the number's cube to end in 1, the last digit of the number must be 1, you can then check the 10's digit - there are only 10.
     
  7. Jul 22, 2004 #6
    Checking the numbers that end in 1 was a good idea... But I don't think Z_100 is a field though (since 100 is composite).
     
  8. Jul 22, 2004 #7

    NateTG

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    I certainly overlooked that issue when I posted, but I'm pretty sure that the set of numbers that are coprime with 100 and zero form a field. Since 11 is coprime with 100 it is still a cubic in a field.
     
  9. Jul 22, 2004 #8
    Ah, I figured you meant the "entire" Z_100.

    *edit* Okay, last nitpick of the day. The set of units modulo 100 won't form a field (3 + 7 doesn't have a multiplicative inverse), however, they will form a group ;)
     
    Last edited: Jul 22, 2004
  10. Jul 22, 2004 #9

    NateTG

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    You are, of course, right. I must be feeling really bright today.
     
  11. Aug 26, 2004 #10
    A shorter answer for this question.

     
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