What is the Probability a Random Number's Cube Ends in 11?

[loseyourname]

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

 

[loseyourname]

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.

 

[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 ;)).

 

[loseyourname]

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 ;)).

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.

 

[NateTG]

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 ;)).

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.

 

[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).

 

[NateTG]

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).

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.

 

[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 ;)

 

[NateTG]

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 ;)

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

 

[supundika]

A shorter answer for this question.

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?

This is actually an easier way and it is shorter than the amount I have written. But here the logical part of it is what solves the problem.

For the last number of the cube to be 1 the last digit of the number which you are calculating the cube should be 1. Which means it should be one pf the following.

x = The last digit is of a cube is 1 (1,11,21,31,41,51,61,71,81,91) = 10 numbers.

So for the second last digit of the cube to be 1 out of those 10 you first of all should get the squared value of those numbers and multiply it by the the number it self. (The example will be useful)

And the combination between the first number of the two digit number with the second last digit of the squared number will give a two digit number. And if the last digit of that number is 1 then the last two digits of the cube are both 1. And there could be only one number like that as each one of them obviously has separate different answers.

eg:- i.) 41*41=1681
...and 1681*41 and [4+8]; 4(first digit of the two gigit no.) + 8
(second last digit of the squared no.) (Here the answer is 12(4+8) So the last digit is 2 so it can't be the answer)

ii.) 71*71=5041
...and 5041*71 and [7+4]; Here the answer is 11. So the last So the second last digit of the cube is 1

So 71 is the only no. between 1 and 100 where the last two digits of its cube are both 1

And if there is only 1 number out of 100 numbers the probability is 1/100 that is 0.01

If you have any doubt about this contact me from the the following e mail supundikadl@yahoo.co.uk