# Enigmatic Code

Programming Enigma Puzzles

## Tantalizer 423b: Body count

From New Scientist #974, 6th November 1975 [link]

The first motion before the conference of Family Doctors was that Miss Emily Scroggins be invited to deliver a lecture on the female epidermis. The Chairman rapped importantly with his gavel:

“I shall put the motion without debate. Those in favour? … Those against? … I declare the motion lost by a majority exactly equal to one quarter of the number voting in favour. Good gracious! Well there’s no need for anyone to be disappointed. Those who wish can view Miss Scroggins tonight at the Golden Tuffet, where she strips to music under the name of Gloria Gunn. What’s that you say, Sir? You would like to change your vote? I daresay you are not alone in that. How many of those previously opposed are now in favour? Twelve, I see. And those previously for but now against? None, I see. This more like it. I declare the motion carried by one vote”.

How many persons were present and voting?

This puzzle and the previous puzzle were both labelled Tantalizer No 423, when originally published in New Scientist. So I’ve labelled this one as 423b to distinguish them.

[tantalizer423b] [tantalizer423]

### One response to “Tantalizer 423b: Body count”

1. Jim Randell 7 August 2019 at 8:59 am

If initially there were F voting for and A voting against. Then “against” has a majority of F/4:

A – F = F/4
4A – 5F = 0

In the second vote 12 of the voters switch from “against” to “for”, and “for” carried the vote by 1.

(F + 12) – (A – 12) = 1
A – F = 23

And we can easily solve these equations to find the total number of voters: A + F.
Run: [ @repl.it ]

```from enigma import matrix, printf

(A, F) = matrix.linear([[4, -5], [1, -1]], [0, 23])

printf("voters = {n} [A={A} F={F}]", n=A + F)
```

Solution: There were 207 voters.
The first vote was 92 votes for, 115 votes against.
The second vote was 104 votes for, 103 votes against.

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