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 $11$ .

#NumberTheory #CheckingCases #PigeonholePrinciple #Modulus #Difference

Easy Math Editor

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`2 \times 3`	$2 \times 3$
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Comments

Indeed. That fact is the motivation behind this problem Brilli the Fortune Teller.

Calvin Lin Staff - 6 years, 9 months ago

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

Michael Diao - 6 years, 9 months 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.

Mehul Gajwani - 6 years, 9 months ago

How do you know that you must have the groups that you listed?

Daniel Liu - 6 years, 9 months ago

Why minus into minus is plus

Harsh Lohia - 6 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$