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# An Interesting Result with Differences

I found this neat little problem on a WOOT handout, and I wanted to share it with you guys.

Given any $$55$$ distinct numbers in the range $$1$$ to $$100$$ inclusive, prove that one can find two numbers with a difference of $$9$$, $$10$$, $$12$$, and $$13$$. In addition, prove that there might not necessarily be two numbers with a difference of $$11$$.

Note by Daniel Liu
3 years ago

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Indeed. That fact is the motivation behind this problem Brilli the Fortune Teller. Staff · 3 years ago

Actually, simply make a chart of all numbers $$\pmod{n}$$ where $$n$$ denotes the number above · 3 years ago

We can look for groups of consecutive numbers such that no two numbers differ by $$9$$, $$10$$, $$11$$, $$12$$ or $$13$$. This will mean that we have groups of $$9$$ numbers separated by $$14$$, i.e.

$$2 - 10;\\ 24 - 32;\\ 46 - 54;\\ 68 - 76; \text{ and}\\ 90 - 98.$$

(You'll see why I didn't start with $$1$$ later.)

However this uses only $$9 \times 5 = 45$$ numbers; we still have to place $$10$$ more.

These will have to be on the side of each group (anything else will lead to differences of $$9$$, $$10$$, $$11$$, $$12$$ and $$13$$, which we do not want). Note however that when we do, it will lead to a difference with two results e.g. if we use $$11$$ that will result in differences of $$9$$ and $$13$$. [The cases of $$99$$ and $$100$$ are insignificant as they only take up $$2$$ out of $$10$$ numbers.]

Using the numbers $$11$$, $$33$$, $$55$$, $$77$$, or $$99$$ (i.e.just after each group) will create differences of $$9$$ and $$13$$. This leaves 5 more numbers.

We can place these before each group (i.e. $$1$$, $$23$$, $$45$$, $$67$$ or $$89$$ ) to then create differences of $$10$$ and $$12$$, without creating a difference of $$11$$.

Hence the final set of numbers will be

$$1 - 11;\\ 23 - 33;\\ 45 - 55;\\ 67 - 77; \text{ and}\\ 89 - 99.$$

========================================================================================== I know this isn't the most rigorous proof but it essentially encapsulates the idea of using the pigeonhole principle. Starting in any other way would fail fairly quickly. · 3 years ago

How do you know that you must have the groups that you listed? · 3 years ago