Given an integer, find the probability its cube ends in 111

[member 428835]

Homework Statement

Given random integer $x$, what's the probability $x^3$ ends in 111?

Relevant Equations

Nothing comes to mind

The last three digits of $x^3$ must be solely dependent on the last 3 digits of $x$. So let $x=a+10b+100c$ for integers $a,b,c$. Then $x^3 = a^3 + 30 a^2 b + 300 a b^2 + 300 a^2 c +O(1000)$ where of course $O(1000)$ don't affect the last 3 digits. Evidently $a^3$ is the only number that governs the 1's digit, and clearly must be $a=1$. Then we have $30b$ being the only value that affects the 10's digit, which implies $b=7$ since $30*7=210$. Thus, the value $14900 + 300 c$ must dictate the 100's digit, which implies $c = 4$ Since $9+3*4 = 21$ which ends in a 1. So this implies of all the numbers, the last 3 digits must be $471$, which is 1/1000 odds, since there are 1000 possibilities from $0-999$ to choose from.

Have I missed something? Sure feels right, but you never know.

 

[PeroK]

Have I missed something? Sure feels right, but you never know.

Check it on a spreadsheet.

 

[member 428835]

Check it on a spreadsheet.

I did, it checks out. But perhaps there are others? I don't think I missed anything htough.

 

[PeroK]

I did, it checks out. But perhaps there are others? I don't think I missed anything htough.

I mean check them all on a spreadsheet.

 

[member 428835]

I mean check them all on a spreadsheet.

No, I did this. Like I tested $i^3 -111 : i = \{1,2,3...1000\}$. Looks like only the one I show above works. I'm more wondering about the logic here: is my thought process correct or have I missed something obvious.

 

[Orodruin]

Homework Statement:: Given random integer $x$, what's the probability $x^3$ ends in 111?

This obviously depends on the probability distribution the integer is drawn from.

 

[PeroK]

No, I did this. Like I tested $i^3 -111 : i = \{1,2,3...1000\}$. Looks like only the one I show above works. I'm more wondering about the logic here: is my thought process correct or have I missed something obvious.

You're not missing anything except why you can't be certain.

 

[vela]

No, I did this. Like I tested $i^3 -111 : i = \{1,2,3...1000\}$. Looks like only the one I show above works. I'm more wondering about the logic here: is my thought process correct or have I missed something obvious.

Looked okay to me.

 

[pbuk]

This obviously depends on the probability distribution the integer is drawn from.

It is true that there is no process for selecting any integer with uniform probability, however in order to answer the question we need to make some reasonable assumption, for example we select among all 10 digit integers, or all positive integers less than $10^n + 1$ for any finite integer $n$, [Edit: added another] or we give the limit for the result as $n \to \infty$ for a uniform selection among all positive integers less then $n$.

 

[Orodruin]

It is true that there is no process for selecting any integer with uniform probability, however in order to answer the question we need to make some reasonable assumption, for example we select among all 10 digit integers, or all positive integers less than $10^n + 1$ for any finite integer $n$.

Exactly, but that is arguably not what is meant by a ”random integer” and is a rather arbitrary assignment. Why not say positive integers smaller than 5783156? That gives a different answer. Why is that a less reasonable assumption? Sure there are a lot of assumptions that will give you the ”expected” answer, but the problem is that the statement does not mention anything about it.

 

[pbuk]

Exactly, but that is arguably not what is meant by a ”random integer” and is a rather arbitrary assignment. Why not say positive integers smaller than 5783156? That gives a different answer. Why is that a less reasonable assumption?

Occam's razor: we make only those assumptions that are necessary to arrive at a consistent result.

Sure there are a lot of assumptions that will give you the ”expected” answer, but the problem is that the statement does not mention anything about it.

Then state your assumption when answering the question. If you are coming to visit me in London and you send me an email asking "how long does it take to get to your place from the airport" I don't answer "that depends on which airport, what mode of transport and what time of day", I respond "its about 30 minutes from Heathrow in a cab, but in rush hour its about twice as long".

 

[Orodruin]

Occam's razor: we make only those assumptions that are necessary to arrive at a consistent result.

Except there is no way to differentiate between different assumptions here. Nothing makes one assumption less extensive than the other. Any assumption that fixes the probability will lead to a consistent result.

 

[pbuk]

Except there is no way to differentiate between different assumptions here. Nothing makes one assumption less extensive than the other. Any assumption that fixes the probability will lead to a consistent result.

Are you implying that there is no way to differentiate between the significance of $f(5783156)$ and $\displaystyle \lim_{n \to \infty} f(n)$? Or that an answer that is meaningful only for one person is as significant as an answer that is meaningful for many people?

 

[Orodruin]

Are you implying that there is no way to differentiate between the significance of f(5783156) and limn→∞f(n)?

Arguably the non-limit is more basic as it does not rely on defining limits. Regardless of what may be a ”reasonable” assumption, a math question should not rely on the students guessing what is being asked.

 

[Office_Shredder]

Orodruin, you're just wrong. Questions like this always refer to the limit.

 

[Orodruin]

Orodruin, you're just wrong. Questions like this always refer to the limit.

Show me the reference.

Or you are saying we should not expect math teachers to pose well defined problems?

 

[PeroK]

Show me the reference.

Or you are saying we should not expect math teachers to pose well defined problems?

As a good Bayesian you could take a uniform distribution on a sufficiently large range of positive integers as your a priori!

 

[Orodruin]

As a good Bayesian you could take a uniform distribution on a sufficiently large range of positive integers as your a priori!

There are many things I could do. Most lead to (at least almost) the same result, but some don’t.

 

[WWGD]

Wouldn't the Chinese Remainder Theorem work here, iirc?
$$ a^3==1(mod 10)==11(mod100)==111(mod 1000)$$?
But then again, I'm not sure the remainder is Chinese ;).
Edit: I mean by tweaking to avoid having 10, 100, 1000 not being pairwise relatively prime?
Edit 2: 471 works.

 

[Frabjous]

Shouldn’t we add a fourth term 1000d with -1<a,b,c<10 and d being any integer greater than -1? Then for every d, there is a 1/1000 chance …
You also need to handle the negative case.