Question

Write the numbers \(1, 2,..., 2n\) on a blackboard, where \(n\)is an odd integer. Pick any two of the numbers,\(j and k\), write \(\left| {j - k} \right|\)on the board and erase\(j and k\). Continue this process until only one integer is written on the board. Prove that this integer must be odd.

Short answer

“The last integer left on the board at the end of the process will to be odd value”.

Step-by-step solution

Introduction

An integer m is said to be an odd integer if it has the following form

\(m = 2n + 1\), for some\(n \in Z\).

Writing given numbers and solving

Write the numbers\(1,2,.....2n\) on a blackboard, where n is to be an odd integer.

The addition of the numbers written on the board will be,

\(1 + 2 + 3 + ..... + x = \frac{{x\left( {x + 1} \right)}}{2}\).

Using the formula above for the addition of first x natural numbers,

\(1 + 2 + 3 + ..... + x = \frac{{x\left( {x + 1} \right)}}{2}\)

The addition defined above is given as follows:

\(\begin{aligned}{}1 + 2 + 3 + ..... + 2n = \frac{{2n\left( {2n + 1} \right)}}{2}\\ = n\left( {2n + 1} \right)\end{aligned}\)

As n and\(2n + 1\)both are odd numbers, the addition of the numbers written on the board\(n\left( {2n + 1} \right)\)will be an odd number.

Now pick any two values j and k from the board.

Erase them and write\(\left| {j - k} \right|\)on the board.

So, the addition of the numbers left on the board is given by

\(n\left( {2n + 1} \right) + \left| {j - k} \right| - j - k = n\left( {2n + 1} \right) + \left| {j - k} \right| - \left( {j + k} \right)\).

Three cases arise

1. One of j and k is an odd value and the other is an even value.
2. Both j and kare even numbers.
3. Both j and k are odd numbers.

Case 1- If only one of j and k is an odd value

Then by property of odd and even values, the addition \(j + k\) and the difference\(\left| {j - k} \right|\)will be odd.

Case 2- If both j and k are even values

Then by property of even numbers, the addition \(j + k\)and then difference \(\left| {j - k} \right|\)will be even.

Case 3- If both j and k are odd numbers

Then by property of odd numbers, the addition \(j + k\)and the difference \(\left| {j - k} \right|\)will be even.

Conclusion

So, if the addition of the numbers left on the board is decreased by an odd number\(j + k\), it is increased by another odd number\(\left| {j - k} \right|\).

And if the sum of numbers left on the board is decreased by an even number\(j + k\), it is increased by another even number\(\left| {j - k} \right|\).

Thus, in both cases, the parity of the addition of the numbers written on the board never changes.

The addition of the numbers\(1,2,.....2n\)written initially on the board is odd; the addition of the numbers left at the end will also be odd.

Therefore, “the last integer left on the board at the end of the process will to be odd value”.