Maths puzzle: How many minions does it take to rig an election?

4 Jul 201623 Shares

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Something we hear over and over again – particularly in relation to how unfair it is that a failing maths grade means a failed Leaving Cert – is, ‘How often do you even use maths in real life, anyway?’

The answer? Constantly.

We use it in our day-to-day life, often without even thinking about it – paying for groceries, planning a budget, calculating how much longer you have to sit idly in front of your computer before you get to leave work…

And, beyond that, maths comes into play in a lot of jobs in a lot of sectors. A lot of them, you’ll know about, like accounting, analytics or teaching.

But have you considered electoral math? Well, you’re about to.

Maths, electoral fraud and minion demographics

In the country Villanya, there are 997,997 people.

Villanya is divided into provinces. Each province is made up of the same number of counties (more than one), each county is made up of the same number of towns (more than one), and everybody in Villanya lives in a town, with the same number of people in every town.

There are some interesting laws in Villanya, too.

If a new person settles in a town, somebody else has to leave, and vice-versa – if someone leaves, somebody else has to come in.

Villanya holds democratic elections, in which everyone is required to vote.

The electoral system is simple. Each town elects a mayor by simple majority (half of the people, plus one). Then all mayors of a county elect a county rep by simple majority. Then all the county reps elect a province governor by simple majority. Finally, all the province governors elect a president by simple majority.

Between 8.4pc and 8.5pc of the population are minions of President Gru, while the rest truly loathe him.

And no wonder. By presidential decree, all minions – and only minions – have a right to reside wherever they like, whenever they like. And, boy, do they like to travel.

But here’s the question. Villanya has an election coming up. Can President Gru get re-elected without committing election fraud?

Scroll down for the solution.

This week’s maths puzzle comes courtesy of Dr Anca Mustata, lecturer in Mathematics at UCC, who is actively involved in the Maths Circles initiative, the Mathematics Enrichment programme in UCC and the Irish Mathematical Olympiad.

Minions goggles

Image via Shutterstock

Solution

So can Gru get re-elected without committing fraud?

Yes, he can!

Let’s suppose that there are seven provinces, each with 11 counties, which each have 13 towns populated with 997 people.

7 x 11 x 13 x 997 = 997,997

Let’s then assume that all minions vote for Gru minions, who are in turn likely to vote for Gru to remain president.

In order to be elected, Gru needs the votes of four provincial governors (out of seven). For argument’s sake, we can assume that the other three governors did not support Gru, and were elected by county officials who did not support Gru. We can therefore discount those three governors in our calculations.

Each of the four Gru-supporting governors needs the votes of six county reps (out of 11). So, in total, Gru needs 24 county reps to be minions.

To be elected, each county rep needs seven of the 13 town mayors to vote for them. That makes a total of 168 minion mayors.

4 x 6 x 7 = 168

Each mayor needs the votes of 499 of the 997 people in a town. This means that, if only minions vote for Gru – and only minions will vote for Gru – he will need 83,832 minions, amounting to 8.400025pc of the population, spread equally among the towns, counties and provinces in order to get elected.

4 x 6 x 7 x 499 = 83,832

As the minions account for between 8.4pc and 8.5pc of the population, this means that, without committing election fraud, Gru can be reelected on the votes of minions alone.

Main image via Narong Jongsirikul/Shutterstock

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Kirsty Tobin is Careers Editor at Siliconrepublic.com, covering careers-related news, features and interviews

editorial@siliconrepublic.com