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5 integers randomly chosen from 1 to 2015. What is the probability that there is a pair of integers whose difference is a multiple of 4?
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Don't know whether it is right or not ? You should check the answer !!


  • If you take a number say 1, then other numbers can be 5,9,13,17,21............ as the diference between these number and 1 is a multiple of 4.
  • Now, if you take 2, then other numbers can be 6,10,14,18,22........... as again difference is a multiple of 4.

Take an example of 5 distinct integers => 102 , 2014 , 517, 343 , 1009 

  • Here, difference of 1009 - 517 = 492 which is offcourse a multiple of 4

Take another example => 555 , 1 , 960 , 1549 ,1968

  • Here, difference of 1968 - 960 = 1008 which is also a multiple of 4.

Take one more example => 129 , 2015 , 17, 740 , 1796

  • Difference between 1796 -  740 is a multiple of 4.

So, if something is divisible by 4, then remainder is 0 and if not, then remainder is either 1,2,3 . 

Now, I guess Dirichlet principle can be applied here also known as pigeonhole principle

You have 5 integer numbers (Pigeons) and 4 labelled blocks (Pigeonholes) . 

Now, take any 5 numbers and see their last digits and divide them by 4 and place them in the particular block , it will surely fall in the blocks labelled as , , 2 , 3 ( These blocks are made based on the remainder obtained by dividing the number by 4 ) 

  • I took this example => 129 , 2015 , 17, 740 , 1796
  • 129 drops into 1 
  • 2015 drops into 3
  • 17 drops into 1
  • 740 drops into 0
  • 1796 drops into 0 . 

Here, by the pigeonhole principle, there is definitely atleast one pair which falls in the same block.

Hence, the probability that there is a pair of integers whose difference is a multiple of 4 will be 1, that is you will always find a pair which belongs to same remainder block.

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4 Comments

Choosing N numbers from a range 1 to R has $\binom{n}{r}$ possibilities, order isnt important.
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@Air1. You are right. Let me think from your example.
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@Air1 and Akriti. Sorry. I am not able to think. I will edit my answer in the comment.
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