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?(adsbygoogle = window.adsbygoogle || []).push({});

[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

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# Is There an Easier Way to Solve This?

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