A Domino Wheel: Upper Secondary Mathematics Competition Question

This is the last of this week’s domino puzzles. As before, take one full set of dominoes and remove the tiles with blanks, leaving you with 21 tiles.

The diagram below shows 8 rectangles placed at the vertices of an octagonal wheel. Each rectangle is to have one domino tile placed inside it. Every tile consists of two squares, each with some pips inside it. The number in the centre of the diagram is the product of the 4 squares taken from the 2 dominoes that are diametrically opposite each other. All four such products must equal the same total and be formed from different tiles from your set of dominoes. Next, orient each tile so that the square with the smallest number of pips is closest to the centre of the diagram, thereby forming a kind of “inner circle”. Calculate the sum of these eight squares, the “inner sum”.

The aim is to find the largest possible “inner sum” of the lowest value squares of the 8 dominoes placed so that the product of the squares of diametrically opposite dominoes is constant.

More Domino Puzzles: Middle Secondary Mathematics Competition Questions

Like yesterday’s Domino Puzzle, you are given a full domino set of 28 tiles. Firstly, remove all the tiles that have a blank end, leaving you with 21 tiles. Now, by looking at each tile as a vertical arrangement of numbers, we are going to use them as fractions. For example, the tile [5,2] can be thought of as the fraction 5/2 or 2/5.

Question A

This time we are going to add together two domino fractions, giving the sum of 6/5. The tile [6/5] is already used as our answer so, using the 20 remaining tiles, how many different sums can you find? The order of the fractions is not important.

Algebraically,

[A/B] + [C/D] = [6/5] = 6/5

where the square brackets [A/B] merely serve as a reminder that these are domino tiles being used as domino fractions.

Domino Puzzle: Lower Secondary Mathematics Competition Question

You are given a full domino set of 28 tiles. Firstly, remove all the tiles that have a blank end, leaving you with 21 tiles. Now, by looking at each tile as a vertical arrangement of numbers, we are going to use them as fractions. For example, the tile [5,2] can be thought of as the fraction 5/2 or 2/5.

Your task is to find all the combinations of 5 distinct tiles such that the product of 4 of the fractions is equal to the 5th fraction, which is itself equal to the number 3.

Algebraically, this equates to:

where the square brackets [A/B] merely serve as a reminder that these are domino tiles. The order of the 4 tiles on the left-hand side is not important.

Short Question

Find all the solutions where the above product equals 3 and that include the tile [5/2].

Project Question

Find all the solutions where the above product equals 3.

Slicing Through Chocolate: Upper Primary Mathematics Competition Question

Imagine you have a bar of chocolate and wish to cut it into individual pieces. Like the chocolate in the image, the bar is one solid piece but moulded to have distinct chunks.

If your bar is 3 chunks wide and 5 chunks long, how many cuts do you need to make so that you have all individual chunks?

A slice means one straight-line cut with a sharp knife. You may not stack pieces on top of, or next to, each other in such a way as to cut through multiple pieces with one cut.

In general, if a bar of chocolate is n chunks wide and m chunks long, how many cuts (as described above) would you need to make in order to separate it into individual pieces?