#iq-test

[online-iqtest]

Can you solve this?

[wiskwa-en]

A `mathematical’ problem

The following `mathematical’ problem is doing the rounds on the internet.

Personally I am not a big fan of this kind of `problem’ because the purpose is not to determine the system behind the data, because there is no unique law that governs it, but whatever it was that was in the head of the poser of the problem.

My standard reply to such a question is: 0 (or sometimes 42, the universal answer from The Hitch-hikers Guide to the Galaxy). Can we say something sensible about a problem like this?

Let us see. For starters: the function of + is clearly not that of addition (if = is supposed to mean equality); this suggested to many people to replace it by some other symbol, @ say. The question then is: is there an operation @ that produces for every pair of natural numbers m and n a natural number m@n and in such a way that 1@4=5, 2@5=12 and 3@6=21? And if yes then what is 5@8?

For a mathematician the answer is clear: apart from the three demands we are free to do as we please, so the answer is “yes” and 5@8 is not determined it can be what we want it to be. For eaxample: define m@n=0, except in the three prescribed cases; that justifies my standard answer. If we separately define 5@8=42 then Deep Thought’s universal answer becomes correct.

Most people do not expect an answer like the two given above but rather a `formula’; some expression with m and n in it that `predicts’ what 5@8 should be. Formulas like that are plentiful. If you look at the data then the three equations are of the form m@(m+3)=x, so it could very well be that the operation simply works with the first coordinate only.

We have three values and these can be fitted by a quadratic polynomial in m alone. You can check that m@n=m2+4m meets our three requirements (plug in m=1, 2, 3). That will produce the answer given by most of the internet: 5@8=45.

With a little twist you can make infinitely many formulas that give the three desired outcomes all produce different results for 5@8. The trick is to find something that produces the value 0 when you substitute 1, 2, or 3 for m and only then. You can take (m-1)(m-2)(m-3). Using that we can produce infinitely many formulas: for every natural number k define m@kn=m2+4m+k(m-1)(m-2)(m-3).

With some tinkering it should be possible to make formulas that have 5@8 take on any value you like; I’d say: get to work. Here’s an example.

It produces 5@8=4754660328285.

To those who want to know if the sequence 5, 12 21 has any `natural’ extensions I recommend the On-line Encyclopedia of Integer Sequences; put the sequence in the search box to see what’s available.

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[wiskwa-en]

A `mathematical’ problem (part 2)

The original version of the previous post elicited a lot of reactions on twitter. This post contains some comments on (and inspired by) these reactions.

1. Some people gave 13 as the answer for 5+8, justifying this by saying that the second and third equalities were false. I think this actually the most sensible solution: if you write `+’ then you must mean `addition’; don’t be silly.

2. The answer suggested most often is 45 and the reasoning is mostly something like the differences are 7 and 9, so going from 3 to 4 it will be 11 and from 4 to 5 it will be 13, and that makes 45 the likely outcome for 5+8.

2a. Alternatively you can try to reconstruct/divine the new meaning of `m+n’; most people came up with `the sum of m*n and m’ (do not write m*n+m because + does no longer mean `add’), this also leads to 5+8=45.

2b. Another answer that came up quite often was 32: that solution simply skips 4+7 and uses 11 to get from 21 to the answer 32.

2c. An other popular answer is 34, the explanation being “you add the true outcome of the addition to the previous answer”. If you were to apply this to the intermediate 4+7 as well then the answer would, again, be 45.

3. In the mathematical magazine Pythagoras, in the issues of April 2009 (The IQ formula) and June 2009 (IQ formula light), you will find an explanation of how to systematically create a formula that describes a given finite sequence of numbers and thus how to find the next term(s) of that sequence. In spite of what the titles of the articles suggest this will not let you necessarily do well on IQ tests because the method does not take any creativity of the makers of the test into account. Oh, and the articles are in Dutch but with some perseverance you will get the mathematical gist of the arguments.

4. The formula from the previous post, the one that gives 5@8=4754660328285, is basically a very contorted (perverse?) application of the method from Pythagoras.

5. In this book (pages 344 and 345) this type of problem is tackled statistically. Some Bayesian arguments are used to determine which of two possible formulas is most probable. But this author also warns that it will never be 100% clear what the poser of the problem was really thinking.