Probability that product of any four natural numbers is divisible by 5

Question

Given any four randomly chosen natural numbers (not mentioned if the numbers taken are distinct or not) what is the probability that their product is divisible by 5?

My answers:

The numbers chosen will either be of the form $5k$ or $5k+1$ or $5k+2$ or $5k+3$ or $5k+4$ ($k$ is a natural number.). Since each of the form is equally likely to occur (I just feel they will be equally likely and don't know the proof) therefore the probability is $1-P(\text{none of the numbers is divisible by 5})=1-(4/5)^4$.

Is my answer along with justification correct?
Now let us consider the question in a different way. Let the product of the numbers be $x$. Since $x$ is equally likely to be of form $5k$ or $5k+1$ or $5k+2$ or $5k+3$ or $5k+4$ (Is it?) therefore answer is $4/5$.

Obviously at least one of the two methods posted answer above is wrong. Which one is it? If both are wrong kindly tell the answer along with justification.

PS the question is from my guide book IIT JEE Mathematics: 35 Years Chapterwise Solved Papers 2013 - 1979

You can't select an element from an infinite set with equal probability. — , Jan 23 '15 at 04:08
The word "randomly chosen" is the key. You need to know what's the "law" under which you are operating. Your second method seems wrong to me, as $x$ need to be equally likely to be of that form, as $x$ is a product of $4$ random numbers, and itself isn't chosen as random. — voldemort, Jan 23 '15 at 04:15
A more well defined question is: Given an interval of $N$ contiguous natural numbers, what fraction of $N \choose 4$ possible distinct products is divisible by 5? — , Jan 23 '15 at 04:18
ok may be you have not got what i am trying to say. consider a square which is painted red. suppose i divide the square in four equal parts. now the red color can be supposed to be made from infinite red spots(number of elements are the infinite spots) but we can easily say that the ptobablity of selecting a red spot from one of the four equal parts of square is 1/4. Hope this helped. I am not commenting about probability of occurence of an element rather i am talking about probability of element occuring in a particular group.@ Eupraxis1981 — humble, Jan 23 '15 at 04:20
Could this question be asking about a uniform choice between the congruence classes modulo $5$? If this is the case, then a uniform distribution is possible, with a probability of $\frac{1}{5}$ for a choice of any one class. — Brian, Jan 23 '15 at 04:21
To practice yourself on an easier problem. Find the probability of the event you describe if the numbers are between 1 and 1000000. Then try the same thing for numbers between 1 and 1000000000. Then from 1 to 10000000000000. You should see a pattern for the probability appearing. What happens if you have an infinity of numbers? — MasB, Jan 23 '15 at 04:21
@BrianScholl Uh i cant understand what you are trying to say i dont know anything about"congruence classes modulo 5" — humble, Jan 23 '15 at 04:25
The version by @BrianScholl seems a lot like what you are asking. Also, your analogy about the square is about an interval while you are asking about sets of individual points. Such sets have zero probability. — , Jan 23 '15 at 04:25
Is this problem quoted more or less verbatim from a book? If so, that is very sad. One can, for fixed $N$, ask for the probability if the $4$ numbers are independently chosen uniformly in the interval $[1,N]$ and take the limit as $N\to\infty$. Then we get the number of your first answer, but strictly speaking the numbers are not randomly chosen from the natural numbers. — André Nicolas, Jan 23 '15 at 04:25
When I say "congruence class modulo 5", I mean to say that you can put each positive integer into one of $5$ categories based on their remainder when divided by $5$. So, you have the class of integers which have remainder $0$, then those with remainder $1$, and so on. This makes it possible to put a uniform distribution that will give the answer you need. — Brian, Jan 23 '15 at 04:30
This question and answer should help: http://math.stackexchange.com/questions/14167/probability-of-picking-a-random-natural-number I believe you need to define a non-uniform probability distribution for the problem to be meaningful. — Thomas Phillips, Jan 23 '15 at 04:31
@Eupraxis1981 link this is the book and is quite popular in india — humble, Jan 23 '15 at 04:31
@BrianScholl yes this is exactly what i mean to say ... thank you — humble, Jan 23 '15 at 04:32
Ok, then your problem makes sense. I noticed that your textbook is a study guide...that's not a bad thing, but its breadth and length suggest that there may be some inaccuracies in the language. — , Jan 23 '15 at 04:36
A good way to ask a question like this on this site would be for the first words of the question itself to be "In Objective Approach to Mathematics, volume 1 (5th edition), which is a study guide for engineering entrance exams popular in India, ... ." People are less likely to see the question as worth answering when they have to dig through a dozen comments to find out this context of the question. — David K, Jan 23 '15 at 04:43
@AndréNicolas: this is the Principle of Spurious Transfinite Induction. "Anything true for all finite N is true for infinity" ;-) — Steve Jessop, Jan 23 '15 at 13:20
I'm trying to create a site for "Indian Competitive Exams" and your question seems to be related to it, I think you would be interested to support it here and spread the word to your friends? — RE60K, Apr 25 '15 at 07:44

score 13 · Accepted Answer · answered Jan 23 '15 at 04:37

13

There are two ways to get the correct solution. The easier way is to consider the set $\left\{\left[5k\right],\left[5k+1\right],\left[5k+2\right],\left[5k+3\right],\left[5k+4\right]\right\}$ of congruence classes modulo $5$ (where $\left[5k\right]$ simply denotes the class with representative of the form $5k$ for an integer $k$, and so on). The probability that you would pick any one of these classes is $\frac{1}{5}$. Thus, the probability that you would pick the class $\left[5k\right]$ is $\frac{1}{5}$. The probability that you will not pick this class is $1-\frac{1}{5}=\frac{4}{5}$. The probability that you would pick $4$ classes that are each not $\left[5k\right]$ is $\left(\frac{4}{5}\right)^{4}=\frac{256}{625}$. Therefore, the probability that you will have a product divisible by $5$ is equivalent to saying that at least one class is $\left[5k\right]$, which has probability $1-\frac{256}{625}=\frac{369}{625}$.

The second method, which is more popular when such classes are not possible, is to take probabilities over the integers in the interval $\left[1,n\right]$, and look at the behavior of these probabilities as $n\to\infty$. Though this is not needed here, it is a good technique to know.

answered Jan 23 '15 at 04:37

Brian

2,147

i get it... can you tell why my second solution is wrong.. and err... why am i getting downvote for this question. i have tried my best to make the question clear. you see that i am in class 10+2 and dont know all the things referred to in comments – humble Jan 23 '15 at 04:40
can you please elaborate the second method you posted in your answer ... how will i look at the behaviour as n→∞ .. (i know limits) – humble Jan 23 '15 at 04:44
1

The reason that your second solution is wrong is that $x$ is not what is chosen randomly, it is the product of $4$ randomly chosen classes. I believe the question has been down voted because the question is ambiguous and ill formed. This is, unfortunately, the fault of the book that you are using. In the future, if you present the problem by saying that it has been copied from a book, then voters who read only long enough to see the question will understand the situation. – Brian Jan 23 '15 at 04:44
1

As for the second method, you would put a uniform distribution on $\left{1,2,\ldots,n\right}$ and find the probability that a randomly selected integers among them is divisible by $5$, then find the probability that a product of $4$ randomly selected integers is divisible by $5$, and then take the limit as $n\to\infty$. This should give the same answer. – Brian Jan 23 '15 at 04:46
@BrianScholl, I looked at your profile and see we are connected. I am from the University of Alabama. Are you a Tigers fan or Tide fan. Looks like you are from south and I suspect that you are a tigers fan. When you are miles apart, we seek some connection without any rivalry to play in. Roll Tide!! – Satish Ramanathan Jan 23 '15 at 07:36
@satishramanathan I am not much into sports, so I can't say that I'm a fan of either. – Brian Jan 23 '15 at 13:17

score 4 · Answer 2 · answered Jan 23 '15 at 04:33

Assuming that you are selecting four numbers from a finite set n which is divisible 4. Then partition that set into four sets.

A = {1,5,9,...,n-3}

B = {2,6,10,....,n-2}

C = {3,7,11,...,n-1}

D = {4,8,12,..., n}

Total number of numbers in each set $= \frac{n}{4}$

In each set, $\frac{n}{5}$ would not be divisible by 5.

The total probability that four numbers picked from these partitions whose product is not divisible by 5 =those numbers picked from each that is not divisible by 5 $= (\frac{4}{5})^4$.

Thus the probability that four numbers picked from these partitions whose product is divisible by 5 $= 1-\frac{256}{625} = \frac{369}{625}$

Probability that product of any four natural numbers is divisible by 5

2 Answers2