2

This is a question that appeared on the 2017 CEMC Galois math contest (which took place a few weeks ago):

A Koeller rectangle:

  • Is a rectangle $m\times n$, n being a whole number with $m \geq 3$ and $n \geq 3$;
  • has parallel lines on its side dividing the rectangle into $1\times 1$ squares;
  • The squares on the edges are white and the ones inside are shaded.

The figure (see attached) is an example of a Koeller rectangle where $m= 8$ and $n= 6$.

Given a Koeller rectangle, r is the ratio between the shaded area and the non shaded area (shaded/non-shaded)

Determine all the possible prime numbers p for which there exists exactly 17 positive values for $u$ for Koeller rectangles with $n=10$ and $r=\frac{u}{p^2}$

Question (in French), and image of example Koeller rectangle:

https://i.stack.imgur.com/x8eC4.jpg

I answered that there are no such primes that satisfy the given conditions, but I'm almost certain that's incorrect. Could someone please explain to me how to solve this question?

Daniel H.
  • 178
  • I have since edited your question as a reference for you, showing the kind of formatting we expect. It may take more time to do so, but you can't expect us to spend time trying to understand your question, *in addition to" providing an appropriate answer. Notice the bullets, notice the separation of text. Now: – amWhy Apr 21 '17 at 16:37
  • @Sqepia It would also be helpful (to you, because it helps people proposing answers) if you at least wrote some of the equations related to the problem. For instance $\frac{8(m-2)}{20+2(m-2)}=r=\frac{u^2}{p}$ is helpful. – Χpẘ Apr 21 '17 at 17:13
  • Thanks for editing and formatting my question, I was kind of in a hurry when I typed it (my lunch hour isn't long enough)... The contest this is from is the CEMC Galois contest (2017) – Daniel H. Apr 21 '17 at 19:04
  • Thanks, Sqepia for filling in some of the details I asked about. That helps a lot. We just like to keep everything "above board" in terms of active, or completed, competitions. I give you a lot of credit for taking time to address this! – amWhy Apr 21 '17 at 20:41
  • Solution here: https://www.cemc.uwaterloo.ca/contests/past_contests/2017/2017GaloisSolution-f.pdf – Jean Marie Feb 17 '22 at 11:27

1 Answers1

2

It is simpler than it looks :)

$$r = \frac{(m-2)(n-2)}{2*(m+n-2)} $$

for n = 10 it gives

$$r = \frac{4*(m-2)}{m+8} $$

$$u = r*p^2 = 4 * p^2 * \frac{m-2}{m+8} = 4 p^2 - \frac{40 p^2}{m+8}$$

The first term is an integer, we need the second one to be so too. So we need M = m+8 to be a divisor of $40p^2$, which itself can be decomposed as follows 2,2,2,5,p,p.

All the possible divisor values for M are

$1,2,4,5,8,10,20,40,$

$p,2p,4p,5p,8p,10p,20p,40p,$

$p^2,2p^2,4p^2,5p^2,8p^2,10p^2,20p^2,40p^2$

That s a total of 24 divisors possible. Remove the ones for which $M<=11$, i.e. $1, 2, 4, 5, 8, 10$, because that would mean $m<=3$. That leaves us with 18 divisors .

For p = 2 and 5 we have lots of duplicates and the total numbers of unique divisors are 6 and 15. For p = 3, we have many values below 11. For primes p above 11 all the 18 divisors are unique (remember p is prime).

The only one that works is p=7, because then the divisors p in the list above disappears (again based on the constraint that M must be >= 11).

In summary

$p=2: 6 divisors $

$p=3: 15 divisors $

$p=5: 10 divisors $

$p=7: 17 divisors $ !!!

any prime p>=11: 18 divisors

Only one solution, p=7

Glad I could help :)

Matt
  • 36
  • 2