The problem feels a bit like the Pigeonhole Principle, but it seems slightly more complex:
Given 12 distinct natural numbers, all of which are less than 37, prove that among the differences between any two of these natural numbers, at least 3 are equal.
My Solution:
Assume that there exists a case where each difference appears at most twice. Let us arrange the numbers in ascending order and discuss the minimal possible sum of their adjacent differences:
The differences would be 1, 2, 3, 4, 5, and 6; among these, 6 appears once, and the others appear twice (this is the minimal case).
Then, the difference between the first number and the last number would be 36: (1+2+3+4+5) \cdot 2 + 6 = 36 In this scenario, the first natural number can only be 0, and this is the only possible distribution for their differences.
However, if we take these 11 differences (1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5) and choose any n numbers to sum up, and then subtract n-1 numbers from within that sum, the resulting differences will include these 11 values again, which would lead to a contradiction with the original assumption.
For example:
The first number is 0
The second number is 5
The third number is 5 + 6
The fourth number is 5 + 6 + 4
Then, (5+6+4) - (5+6) = 4. Doesn’t this mean the difference 4 has appeared 3 times?This shows that, in any case, a situation where only two differences are equal cannot exist.
By the way, let’s find someone born on the same year, month, and day as you:
Check how many people are in your grade at school. If there are more than 500, and you are not some special person, there will definitely be someone born on the same year, month, and day as you!
Do you know why? This isn’t some “destiny”!
There is a mathematical basis for this. Why not analyze it?
If you don’t believe it, ask around in every
class (^_^)? You will definitely find someone.
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