# Thread: Probability involving a number cubed and its last two digits

1. ## 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

2. Originally Posted by Shapeshift
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) ....

3. Originally Posted by Shapeshift
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