As we can see that the favourable case are those where sum of pair of two natural numbers is divisible by 4, and such pairs may be listed as below:

\begin{align*} &(0,0), (0,4), (0,8),\\ &(1,3), (1,7),\\ &(2,2), (2,6),\\ &(3,1), (3,5), (3,9)\\ &(4,0), (4,4), (4,8)\\ &(5,3), (5,7)\\ &(6,2), (6,6)\\ &(7,1), (7,5), (7,9),\\ &(8,0), (8,4), (8,8),\\ &(9,3), (9,7). \end{align*} Thus, there are 25 favourable cases where sum of two natural numbers is divisible by 4. Now, for exhaustive cases, we can see that there total 10 natural number viz. 0--9. Therefore, there are 10$\cdot$10=\,100.

Therefore, the probability comes out to be \begin{align*} P=\,\frac{25}{100}=\frac{1}{4}. \end{align*}

Is my approach is correct?

  • 6
    $\begingroup$ First of all, by natural numbers you mean numbers $0,1,2,\dots,9$? Because $10,11, \dots$ etc. are also natural numbers. Second, no, you should not ignore $9$ from the possible cases. These are "all possible cases" not "potentially favorable cases". $\endgroup$
    – Jimmy R.
    Apr 4, 2018 at 12:43
  • 3
    $\begingroup$ What about $(9,3)$ (and quite a few more)? $\endgroup$ Apr 4, 2018 at 12:44
  • 2
    $\begingroup$ The sum has to be less than 10? $\endgroup$
    – Ripstein
    Apr 4, 2018 at 12:45
  • 1
    $\begingroup$ The sum should be divisible by 4 only $\endgroup$
    – IgotiT
    Apr 4, 2018 at 12:45
  • 1
    $\begingroup$ Should $(9,0)$ really be included? $\endgroup$
    – Arthur
    Apr 4, 2018 at 13:02

2 Answers 2


You can also easily generalize this to all natural numbers. A number has a certain remainder, when divided by 4, with those numbers divisible by 4 having a zero-remainder.

It is given, that if you add numbers, and calculate this remainder, you can simply add the remainders. So for 2 numbers, having the remainder of 2 when divided by 4, their sum will have a remainder of 4, which is essentially the same as zero.

You choose two numbers randomly and you can basically just choose their remainder, since this is all that matters. You have the possibilites $0,1,2,3$, for which you gave an exhausitve list in your question. There is one possible remainder for the second number, regardless of what number is our first. Since the remainders are equally likely, you can see, you have a 1 in 4 chance of having a number divisble by 4, if you sum two randum numbers.

  • $\begingroup$ "the remainders are equally likely" On what is this assertion based? $\endgroup$
    – drhab
    Apr 4, 2018 at 14:28
  • $\begingroup$ How do you choose a natural number? There is no uniform distribution on $\mathbb N$. $\endgroup$ Apr 4, 2018 at 17:07
  • $\begingroup$ I assumed, that you choose each number with equali propability. Since there is no distribution given in the question, I assumed it to be discrelty uniform. $\endgroup$
    – Laray
    Apr 5, 2018 at 6:24

The question is easier to answer if the number of allowed numbers is a multiple of $4$. In this case, whatever the first number is, exactly $\frac{1}{4}$ of the numbers have the correct residue modulo $4$ to give a sum divisible by $4$ , hence the answer is $\frac{1}{4}$.

Your approach is correct (Laplacian case)

  • $\begingroup$ Can you please tell how you come to the conclusion that exactly 1/4 of numbers have the correct residue modulo 4? $\endgroup$
    – IgotiT
    Apr 4, 2018 at 16:05
  • $\begingroup$ If we fix one number (and assume that we have $4k$ allowed numbers), then the residue of the sum modulo $4$ depends only on the residue of the second number modulo $4$. Hence, each residue occurs exactly $k$ times. This approach does not work in our case, because $10$ numbers are allowed. $\endgroup$
    – Peter
    Apr 4, 2018 at 17:08

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