IQ test: weighing the coins

By Frederic Friedel

Many years ago, when I was in my late teens, and egged on by friends, I decided I was smart enough to apply for Mensa membership. Mensa is the largest and oldest high IQ club in the world, reserved for people who score at the 98th percentile or higher on a standardized, supervised IQ or other approved intelligence test. The word mensa means “table” in Latin, as seen in the organization’s round-table logo. It symbolizes the coming together of equals.

Well, I contacted the German chapter of Mensa, which today looks like this. After some months I was invited to take an IQ test at one of the centers, of which there were a handful at the time (today there are over 80). I sat for the test with a couple of other people, handed in the papers, had a snack with a some very nice wannabe geniuses, and left.

For membership Mensa required an IQ of over 130 — using a test where the mean score for the population is 100 and genius is above 140. Today there are IQ tests where people score over 200, and certainly it is common to hear hype about people who have astronomical IQs. Two of my best friends, Judit Polgar (stongest female chess player in history, “IQ 170”) and Garry Kasparov (legendary World Chess Champion, “IQ 194”) are in this top ten list, which ranges from 160 (Stephen Hawking) through Einstein (IQ 160–190), da Vinci (180–190) to somedude with an IQ of 250–300. But that is a different world — of hype. In my Mensa IQ test the top score was in the 140s.

Well, after a couple of weeks I received a phone call asking me to come to the Mensa chapter for an interview. I could tell by the tone of the caller’s voice that something dramatic was in the air. And when I turned up there were a number of people waiting, in silent reverence, The reason: I had scored higher than anyone in the Club’s history. In the 170s.

Unfortunately, like Dorcus Lane in Candleford, I have one weakness: I am quite honest. So I didn’t last long at the Mensa center. Instead of enjoying protracted adulation from the officers and members, I confessed: it was a scam. I had practiced IQ tests, intensely, for a month. The test they were using was a standardized one, and I had got hold of a bunch of them. They were all variations of the same pattern, similar to the ones you will find if you google “IQ test” today. What I had proved — and this was taken into account in the subsequent Mensa testing — was that you can learn to score high in these tests, by simply practicing.

I wasn’t the most popular person at the Mensa chapter after that, but I did get invited to a couple of their Annual Gatherings, with members of the local and international clubs. There were lectures, workshops, games and outings — all in the spirit of Mensa. This is how it worked:

On one outing we all rode through a lovely forest on bicycles. Before we started we were given a logical puzzle, after which we mounted the bikes and rode at a fair speed on the forest path. When anyone thought they had the solution he or she would ring their bell and everyone would stop to listen. If the problem was solved we would get another one and proceed on the ride.

There was one problem which was particularly difficult, and that is the point of this article. You have twelve coins, one of which is slightly lighter or heavier than the other eleven. You have a pair of scales like the ones shown on the left. Your task is to find the defective coin in three weighings. It is not trivially easy — and a warning: I am not going to give you the solution in the end!

Start thinking: if you put six coins in each pan, one will go up and the other down. But all you have done is to determine that the defective coin might be lighter and in the left pan or heavier and in the right. How do you proceed?

If you divide the coins into three groups you have four coins in each, and if you make four groups you have three in each. You can start with a 4–4 weighing, and if the scale balances the bad coin has to be in the third group. But if it doesn’t you have a similar problem: it could be lighter and in the one pan or heavier and in the other. Depending on the result of a weighing you decide how to continue. For convenience name the coins 1–12, so a weighing can be recorded as 1–2–3–4 against 5–6–7–8, and 1–2–3–4 against 9–10–11–12 would be a second weighing. Of course removing coins counts as a second weighing — e.g. 1–2–3 against 5–6–7 after the first weighing above.

Note that you must identify the defective coin, but not necessarily know if it is lighter or heavier. A couple of people have found the problem impossible to solve because they were trying to establish, sine qua non, the latter.

Well, that’s all the help you are going to get. I need to mention that it was me who rang the bell in the forest, and I received commendation for solving the problem (which was completely new at the time). Since then I have given it to a large number of people, even published it in a magazine I was editing. On that occasion my friend John Nunn from England wrote in with an annoying corollary: find the defective coin while specifying all three weighings in advance. If you were able to solve the base problem above — where you can proceed with each weighing depending on the result of the previous one — turn your attention to John’s problem. That should really give you something to think about.