Question

Given 11 different integers from 1 to 20, prove that at least two of them are exactly 5 apart.

Answer 1

In the group of integers from 1-20, we will have 10 even and 10 odd integers.

Let's assume we select the 10 even integers first. Now, none of these numbers will have an odd difference between them as all of them are an even difference apart.

(To understand better, consider the numbers 2,4,6,8. Now, the difference between these numbers is even, no matter which two numbers are considered. So, we can never get 5 or any other odd number as the difference here.)

However, since we have selected all 10 even integers, the 11^th number has to be one of the odd integers. When an odd integer is introduced along with a set of even integers, we will get an odd number difference each time we consider the odd integer with an even integer.

(To understand better, consider the numbers 3,4,6,8. Now, each time we consider 3 along with an even number, we will get an odd number difference. Consider 3,6 for instance, the difference is 3. Consider 3,8 now, the difference is 5.)

So, we can generate an odd number difference as long as we have atleast one odd integer in a set of even integers. The same logic holds true for atleast one even integer in a set of odd integers. Since we are selecting 11 numbers, we have to inevitably choose atleast one odd (even) integer after selecting all 10 even (or odd) integers and so, we can very well generate an odd number difference (in this case, 5) between atleast two of the selected numbers. Hence, proved.

(Kindly note that I've considered the extreme cases (all odd or all even) to prove the required statement. As the extreme cases hold true, all other cases being a combination of these extreme cases, will hold true as a result.)

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Hope this helps!