IN HONOUR OF Maths Week, as is our annual tradition, we’re setting our readers some puzzles. Give them a go!
In October 1843, William Rowan Hamilton invented a new type of algebra called quaternions. The story is well known to mathematicians around the world and the annual Hamilton Walk organised by Maynooth University commemorates this event by the banks of the Royal Canal at Broombridge, Dublin.
Hamilton also created a game where the challenge is to find a route around a playing board that visits 20 cities once and only once. This is a branch of maths called Graph Theory.
A graph in this context being a diagram that shows the connections between different points. The connections could for example be roads and the points, cities or perhaps social media connections and people.
A journey that can travel around a graph visiting each point once and only once is called a Hamiltonian Path. If it starts and finishes at the same point, it is called a Hamiltonian Cycle.
Today’s puzzles are about Hamiltonian Cycles.
1. As a warm up, see if you can find a Hamiltonian Cycle on the following graph.
2. A snake-oil salesman travels from town to town pedalling his dodgy wares. He must be
careful to plan his journey and visit as many towns as possible but also making sure not to visit the same town twice or he could be attacked by dissatisfied customers. There are 64 towns within easy reach of each other.
Having some knowledge of graph theory, he realises that the 64 towns can be represented by the squares on his chessboard and his journeys between them represented by a rook’s moves. That is he can move horizontally and vertically in straight lines and must be careful not to pass through the same square twice.
He also remembers that in his market research, he already tried two towns and they definitely would not welcome him back. His starting point is his hometown, marked “S” and the two towns he must avoid are each marked with an X.
Can you find a route that will start at S and (moving in straight vertical and horizontal lines) visit each square once and only once (except those marked with an X) and return to S?
Wednesday’s puzzles: The answers
1. I buy 10 pieces of fruit for €10. The apples cost 80c each and oranges cost €1.30 each.
How many apples and how many oranges did I buy?
Answer: 6 apples and 4 oranges.
We will go through the algebra step by step.
We know that the number of apples and the number of oranges add up to 10.
If we use the symbol Ap for the number of apples and Or for the number of oranges, we can write an equation for this information.
Ap + Or = 10
We also know that the amount spent is €10 and the price of an apple is €0.8 and the price of orange is €1.30.
The total spent on apples is Ap x €0.80. The total spent on orange is Or x €1.30. These both add up to €10.
We can write this as an equation.
0.8Ap + 1.3Or = 10
We know already that Ap + Or = 10.
We subtract Or from both sides of the equation to get an expression in terms of the number of apples:
Ap = 10- Or
Our other statement is 0.8Ap + 1.3Or = 10
And we can replace the Ap in this equation with 10 – Or (as they are equal)
And we get the expression: 0.8(10 – Or) + 1.3Or = 10
Multiplying this out gets us 0.8(10) – 0.8Or + 1.3Or = 10
Rearranging:
8 +1.3Or – 0.8Or = 10
8 + (1.3 – 0.8)Or = 10
8 + 0.5Or = 10
Subtracting 8 from both sides gives:
0.5Or = 2
Multiplying both sides by 2 gives: Or = 4
Therefore, the number of oranges is 4 and the number of apples must be 6.
Check to make sure this makes sense:
4 x 1.30 + 6 x 0.8 = 5.20 + 4.80 = €10
Which shows the answer is correct.
Come back tomorrow for the answers to today’s puzzle.
The puzzles this week have been compiled for The Journal by Eoin Gill of Maths Week Ireland and South-East Technological University (SETU).
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