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 \(11\).

## Comments

Sort by:

TopNewestIndeed. That fact is the motivation behind this problem Brilli the Fortune Teller. – Calvin Lin Staff · 2 years, 10 months ago

Log in to reply

Actually, simply make a chart of all numbers \(\pmod{n}\) where \(n\) denotes the number above – Michael Diao · 2 years, 10 months ago

Log in to reply

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 · 2 years, 10 months ago

Log in to reply

– Daniel Liu · 2 years, 10 months ago

How do you know that you must have the groups that you listed?Log in to reply

Why minus into minus is plus – Harsh Lohia · 2 years, 10 months ago

Log in to reply