# Is There an Easier Way to Solve This?

1. Jul 21, 2004

### loseyourname

Staff Emeritus
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?

$$n = (1, 2, . . ., 1000)$$

$$n = x + y, \ x = (100, 200, . . ., 900), \ y = (1, 2, . . ., 100)$$

$$n^3 = (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$$ =

$$(x^3 + 3x^2y + 3xy^2) + y^3$$ =

$$[x(x^2 + 3xy + 3y^2)] + y^3$$

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

Let $$z = x^2 + 3xy + 3y^2$$

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

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

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

$$y^3 = 100a + 11$$

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

Clearly there is only one solution to this equation, so $N(A)$ = 1, so

$$P(A) = \frac{N(A)}{N}$$ = 0.01

Last edited: Jul 21, 2004
2. Jul 21, 2004

### loseyourname

Staff Emeritus
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.

3. Jul 22, 2004

### Muzza

A lot of what you've done can be simplified if you use modulo arithmetic. You want to find the number of solutions to $$x^3 \equiv 11 \pmod{100}$$ where $$1 \leq x \leq 100$$. Suppose $$x$$ is such a number that fulfils both equations. Then (by the very definition of modulo) there is an integer $$k$$ such that $$x^3 = 100k + 11$$, 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 ;)).

4. Jul 22, 2004

### loseyourname

Staff Emeritus
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.

5. Jul 22, 2004

### NateTG

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):
$$0^3=0$$
$$1^3=1$$
$$2^3=8$$
$$3^3=27$$
$$4^3=64$$
$$5^3=125$$
$$6^3=216$$
$$7^3=343$$
$$8^3=512$$
$$9^3=729$$
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.

6. Jul 22, 2004

### Muzza

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).

7. Jul 22, 2004

### NateTG

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.

8. Jul 22, 2004

### Muzza

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
9. Jul 22, 2004

### NateTG

You are, of course, right. I must be feeling really bright today.

10. Aug 26, 2004

### supundika

A shorter answer for this question.