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

• member 428835
This could be a reasonable assumption for a mathematical problem, but it may not be a realistic assumption for a real-world situation. It ultimately depends on the context in which the question is being asked.
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.

joshmccraney said:
Have I missed something? Sure feels right, but you never know.
Check it on a spreadsheet.

PeroK said:
Check it on a spreadsheet.
I did, it checks out. But perhaps there are others? I don't think I missed anything htough.

joshmccraney said:
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.

PeroK said:
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.

joshmccraney said:
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.

pbuk
joshmccraney said:
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.

joshmccraney said:
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.

Orodruin said:
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 ##.

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pbuk said:
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.

Orodruin said:
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.

Orodruin said:
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".

pbuk said:
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.

Orodruin said:
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?

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pbuk said:
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.

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

Office_Shredder said:
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?

Orodruin said:
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!

PeroK said:
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.

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.

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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.

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## 1. What is the probability of an integer's cube ending in 111?

The probability of an integer's cube ending in 111 is 1/100, as there are 100 possible outcomes for the last three digits of a cube (000, 001, 008, 027, 064, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 24389, 27000, 29791, 32768, 35937, 39304, 42875, 46656, 50653, 54872, 59319, 64000, 68921, 74088, 79507, 85184, 91125, 97336, 103823, 110592, 117649, 125000, 132651, 140608, 148877, 157464, 166375, 175616, 185193, 195112, 205379, 216000, 226981, 238328, 250047, 262144, 274625, 287496, 300763, 314432, 328509, 343000, 357911, 373248, 389017, 405224, 421875, 438976, 456533, 474552, 493039, 512000, 531441, 551368, 571787, 592704, 614125, 636056, 658503, 681472, 704969, 729000, 753571, 778688, 804357, 830584, 857375, 884736, 912673, 941192, 970299, 1000000).

## 2. How can the probability of an integer's cube ending in 111 be calculated?

The probability of an integer's cube ending in 111 can be calculated by dividing the number of integers whose cube ends in 111 by the total number of possible outcomes for the last three digits of a cube (100). This can also be expressed as 1/100 or 0.01.

## 3. What is the significance of an integer's cube ending in 111?

An integer's cube ending in 111 has no particular significance, as it is just one of the 100 possible outcomes for the last three digits of a cube. It does not have any special mathematical properties or implications.

## 4. Are there any patterns or relationships between an integer and its cube ending in 111?

There are no known patterns or relationships between an integer and its cube ending in 111. The occurrence of an integer's cube ending in 111 is purely random and cannot be predicted based on the integer itself.

## 5. Can the probability of an integer's cube ending in 111 be increased or decreased?

No, the probability of an integer's cube ending in 111 cannot be increased or decreased. It will always be 1/100 or 0.01, regardless of the integer being considered. This is because the last three digits of a cube are determined solely by the last three digits of the original integer, and there is no way to manipulate those digits to specifically result in 111.

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