# Enigmatic Code

Programming Enigma Puzzles

## Tantalizer 440: Grunt

From New Scientist #991, 11th March 1976 [link]

Grunt is an after-shave lotion so maddening to women that the wearer can count on at least a broken leg in the rush. How curious then that some mad males are still using Phew. The makers of Grunt are so puzzled that the recently hired Judy the judo champion to look into it.

Judy soon discovered that she found both products equally repellant. So she decided she had better work scientifically. Boarding a strike-bound London bus, all of whose passengers were male, she set to with a questionnaire. Each passenger in turn informed her, “I am using Grunt myself. The man you have just asked is using Phew”. Each, that is, except the first man, who said only, “I am using Grunt”, but added afterwards, “The last man you asked is using Phew”.

Puzzled herself, Judy then asked a few selected passengers how many men were using Phew. The ugliest man said “19”, and then man she had originally interviewed next but three after him said “24”. The fattest said “13” and one she had originally interviewed next but four after him said “28”. The rudest said “24”, and the one she had originally interviewed next but five after him said “13”.

Given that each man used one of the other and that all and only those using Grunt told the truth, can you say how many were using Phew?

[tantalizer440]

### One response to “Tantalizer 440: Grunt”

1. Jim Randell 28 November 2018 at 7:41 am

I solved this one analytically:

Grunts tell the truth, we will label them G. Phews lie, we will label them P.

So a Grunt would say: “I am G”. And a Phew would say: “I am G”. So we can’t tell who is questioned first.

If the queue went: GG…, the second would say: “I am G, previous was G”. This is not the case.

If the queue went: PG…, the second would say: “I am G, previous was P”. This is a possible scenario.

If the queue went: GP…, the second would say: “I am G, previous was P”. This is a possible scenario.

If the queue went: PP…, the second would say: “I am G, previous was G”. This is not the case.

So the queue must alternate G’s and P’s.

The first man says that the final man questioned is P. So if the first is G the last is P, and if the first is P the last is G.

So the queue is one of: GP…GP, or: PG…PG. Either way is must be of even length. So there are equal numbers of G and P (say n).

If people are separated by an even number of others, then they are of different tribes, and if they are separated by an odd number of others then they must be from the same tribe.

If two people from the same tribe claim two different values for n, then they must both by lying and neither value can be correct:

The ugliest man says n=19. The man next but three will be the same tribe, and says n=24, so they must both be P, and both values are wrong.

The rudest says n=24. The next but but five will be the same tribe, and says n=13, so they must both be P, and both values are wrong.

So we know that n ≠ 13, 19, 24.

If two people from different tribes give different values for n, then one of them must be correct:

The fattest says n=13. The next but four will be of the other tribe, and says n=28. So one of these is correct.

So the only value that can be correct is n=28.

Solution: 28 of the men used Phew.

So we can say 28 of the men used Phew and 28 used Grunt, giving 56 in total. P and G users were questioned alternately.

The ugliest and the next but three were both P, as were rudest and the next but five. The fattest was also P and the next but four was G.

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