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The latest posts discuss problems in the recent Fryer / Galois Contest (for Grades 9&10 students). Those problems and their solution process are really interesting.

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Shading Fraction of a Rectangular and Koeller Ratio – Flavour of Galois 2017

Another interesting problem in Galois 2017 is the following one, which considers the Koeller Shading Ratio. The problem is as below.

Problem. A Koeller rectangle:

  • is an m by n rectangle where m,n are integers with m>=3 and n>=3,
  • has lines drawn parallel to its sides to divide it into 1 by 1 squares, and
  • has the 1 by 1 squares along its sides unshaded and the 1 by 1 squares that do not touch its sides shaded (as in figure).

An example of a Koeller rectangle with m = 6 and n = 4 is shown. (The cell marked S is shaded, the blank cell is unshaded.)

 S S S S
 S S S S


For a given Koeller rectangle, let r be the ratio of the shaded area to the unshaded area.

Now are the questions:

(a) [Answer only] Determine the value of r for a Koeller-rectangle with m = 14 and n = 10.

(b) [Full solution required] Determine all possible positive integer values of u for which there exists a Koeller-rectangle with n = 4 and r = (u/77).

Well, readers are advised to try first before reading the solution below, followed by some remarks.


First, develop some formulae.

Unshaded region = m + m + n + n – 4 = 2 (n+m-2)

Shaded region = (m-2) (n-2) = mn – 2m – 2n +4

Ratio r (of shaded to unshaded)

r = (m-2) (n-2) ⁄ (2 (n+m-2))

For (a), m = 14, n = 10, therefore

r = (12) (8) ⁄ (2 (22)) = (24) ⁄ (11) .

For (b), since n = 4, r = (m-2) ⁄ (m+2)

As its said that r = (u / 77), so:

u = 77 r = (77) (m-2) ⁄ (m+2)  (*)

As (m-2) ⁄ (m+2) = 1 – [4 ⁄ (m+2)], only 2 or 4 divides both m+2 and 4 (besides 1 as trivial factor).

In order that u be an integer, m+2 must contains a factor of 77, as well a factor 2 or 4.

As 77 = 7 x 11, so we will consider the cases

(a) m+2 = 7, 14, 28 –> m = 5, u = 33; m = 12, u = 55; m = 26, u = 66;

(b) m+2 = 11, 22, 44 –> m = 9, u = 49; m = 20, u = 63; m = 42, u = 70;

(c) m+2 = 77, 154, 308 –> m = 75, u = 73; m = 152, u = 75; m = 306, u = 76.

Therefore, possible values of u are 33, 55, 66, 49, 63, 70, 73, 75, 76 (total nine solutions).


(1) For Koeller shading ratio r, we recommend to work with formula from the very beginning; it is clearer and reduce repetitious work. The formula links r with n, m.

(2) In question (b), given 77 as the common denominator of a series of ratios, it asks to find all possible integers u to appear in the numerator – this requires an analysis of factors – as in reduction of fractions.

(3) As u is an integer, an explicit expression of u – as in (*) — is convenient to work with.

Finally, note that has 3 x 2 x 2 = 12 factors. But m+2 cannot be 1, 2 or 4 (as m>=3). —

Therefore, the other 12 – 3 = 9 factors correspond to the nine solutions.

This is just a step-by-step work with algebra, plus some analysis of factors. A neat and clear presentation is what is asked by the problem.

Some Group Paid Special Priced Breakfast– Flavour of Galois 2017

In this year’s Galois math contest, one problem about the special-pricing day breakfast is fairly interesting. Here we take questions (b) and (d) of this problem to the readers for a glimpse into, – and even to taste its flavour.

The Problem.

The Breakfast Restaurant has a special pricing day. If a customer arrives at the restaurant between 4:30 a.m. and 7:00 a.m., the time that they arrive in hours and minutes becomes the price that they pay in dollars and cents.

For example, if a customer arrives at 5:23 a.m., they will pay $5.23.

(b) [Answer only] Robert arrived 10 minutes before Emily, and both arrive during the period of the special pricing. In total, they paid $12.34. What were their arrival times?

(d) [Full Solution Required] Larry and Mio arrived separately during the period of the special pricing. In total, they paid $11.98. Determine the range of times during which Larry could have arrived.

The problem is so easy as everyone can understand it, right? – But it takes some thinking to get the solution correct and perfect! By the way, when it’s suggested “Full solution required”, students will have to write not only the final answer, but also solution steps.

If we skim forward to questions (b) and (d), we find a group of two persons come to the restaurant, and it’s given of the price they paid together; and then students (writing the test) is asked to deduct about amount of each paid as well as arriving times according to some more given conditions. Not in a hurry for details, one may ask self (as a warm-up) that what total price paid is possible (valid)?

The price for two people must be between $8.60 (if both arrived at 4:30 am) and $14.00. But some price for example, $13.30 is not possible. Suppose both arrived at 6: 59, they paid $6.59 each, so total of $13.18; if one of them arrived (just one minute late) at 7:00, the total paid shall became $13.59. If both arrived $7.00, that’s $14.00 total. – Obviously a jump! It’s caused by 1 hour = 60 min however $1 = 100 cents.

For question (b) it’s straight and we can separate hours with minutes. The hour they arrived? Just divide 12 by 2 to get 6. What’s the minutes they arrived? It’s like finding two integers with sum 34 and difference 10. You got it? Robert arrived at 6:12 am and Emily did 10 min later at 6:22.

Any other possibilities? Well no. We know we will convert 100 cents to $1. With the total price given as 12.34 cents, is it possible that the cents they should pay (according to the minutes they arrive) add up to 134 cents? Not possible, since the maximum is 59 + 59 = 118 (We’ve done that!). — The thinking does not yield new solution but as many know, contest questions may be tricky so be wary.

For question (d), the total cents paid were 98 cents (NOT possible to be 198 cents), so we are sure completely that the hours they arrived add to 11 (= 5+6), and the minutes they arrived add to 98. Decomposing 98 as the sum of two integers, while both being between 0 and 59, we have:

98 = 39 + 59 = 40 + 58 = … = 59 + 39

But here indeed there’s a tricky stuff:
it’s said in the question that Larry and Mio arrived separately, no mentioning about who were earlier.

The one arrived early was after 5:00, and the late comer was after 6:00. Therefore, Larry could have arrived between 5:39 to 5:59 or between 6:39 to 6:59.

Worthy to note that it’s not possible for any of the two to arrive between 6:00 to 6:38.

With some additional effort, one shall be able to present these steps of solution nicely.