- #1

Hill

Gold Member

- 704

- 557

- Homework Statement
- A gardener collected 17 apples...

He finds that each time he removes an apple from his harvest, he can split the remaining apples in two piles of equal weight, each containing 8 apples. Show that all apples are the same weight.

- Relevant Equations
- Linear algebra

In fact, it WAS a homework couple of years ago, and I've solved it, kind of (below). I still would like to find a cleaner solution.

Here is what I did.

Let's say, the apples are labeled, and their weights are ##x_1, x_2, ...##. He takes out the apple #1 and finds that, e.g., ##x_2+x_5+x_9+... = x_3+x_4+x_6+...##. Then he puts the #1 back, takes out #2, and finds that ##x_1+x_7+...=x_3+x_8+...##. And so, 17 times. Each side of each equation has 8 apple weights in it. We are asked to show that ##x_1=x_2=x_3=...##.

In terms of linear algebra, the problem can be expressed as the matrix equation, ##A x = 0##, where ##x## is a vector of ##x_1, x_2, ... , x_{17}##, ##A## is a 17x17 matrix with 0 along the diagonal, and +1 or –1 at the other positions such that each row has eight +1s and eight –1s, ##0## is the 17-component zero vector.

From the equation, ##x## is the null space (the kernel) of A. The problem is to show that for all matrices ##A## of this form, the null space has rank 1. For this, rank of A has to be 16. That is, if we take a 16x16 minor matrix of A, its determinant is not 0. So, we take a 16x16 matrix with 0s on diagonal and 1s and -1s everywhere else and consider its determinant.

The determinant is sum of products of all possible permutations of matrix elements, one from each row and from each column, with corresponding coefficients 1 or -1.

The terms that include 0s from the diagonal are 0s and don't contribute to the sum. Each row has 15 non-zeroes. There are 16 rows. So, there are 15^16 combinations of 1s or -1s. This is an odd number.

For every two rows, 14 non-zero pairs are from a same column. These do not contribute to the determinant. There are 14 times a number of pairs of rows of such permutations, which is an even number.

After removing the latter even number from the odd number above (i.e. from 15^16) we are left with some odd number of terms that contribute to the determinant. Each term is equal to either 1 or -1 and contributes with a coefficient of either 1 or -1. Thus, the determinant is a sum of odd number of 1s and -1s.

Here comes the punch line:

So, determinant of this 16x16 matrix is not 0. QED.

But there is some handwaving there, isn't it? Is there a better way to solve it?

Here is what I did.

Let's say, the apples are labeled, and their weights are ##x_1, x_2, ...##. He takes out the apple #1 and finds that, e.g., ##x_2+x_5+x_9+... = x_3+x_4+x_6+...##. Then he puts the #1 back, takes out #2, and finds that ##x_1+x_7+...=x_3+x_8+...##. And so, 17 times. Each side of each equation has 8 apple weights in it. We are asked to show that ##x_1=x_2=x_3=...##.

In terms of linear algebra, the problem can be expressed as the matrix equation, ##A x = 0##, where ##x## is a vector of ##x_1, x_2, ... , x_{17}##, ##A## is a 17x17 matrix with 0 along the diagonal, and +1 or –1 at the other positions such that each row has eight +1s and eight –1s, ##0## is the 17-component zero vector.

From the equation, ##x## is the null space (the kernel) of A. The problem is to show that for all matrices ##A## of this form, the null space has rank 1. For this, rank of A has to be 16. That is, if we take a 16x16 minor matrix of A, its determinant is not 0. So, we take a 16x16 matrix with 0s on diagonal and 1s and -1s everywhere else and consider its determinant.

The determinant is sum of products of all possible permutations of matrix elements, one from each row and from each column, with corresponding coefficients 1 or -1.

The terms that include 0s from the diagonal are 0s and don't contribute to the sum. Each row has 15 non-zeroes. There are 16 rows. So, there are 15^16 combinations of 1s or -1s. This is an odd number.

For every two rows, 14 non-zero pairs are from a same column. These do not contribute to the determinant. There are 14 times a number of pairs of rows of such permutations, which is an even number.

After removing the latter even number from the odd number above (i.e. from 15^16) we are left with some odd number of terms that contribute to the determinant. Each term is equal to either 1 or -1 and contributes with a coefficient of either 1 or -1. Thus, the determinant is a sum of odd number of 1s and -1s.

Here comes the punch line:

**No sum of odd number of 1s and -1s makes 0 !!!**So, determinant of this 16x16 matrix is not 0. QED.

But there is some handwaving there, isn't it? Is there a better way to solve it?