The answer to those broken arithmetic IQ test posts is there is no answer

mathematics - algebra, algorithms, IQ, IQ test, perl

I happened to be looking around on twitter recently (bear in mind this doesn't happen often) and I came across the following:

This kind of thing make me mutter to myself and shake my head. Because in fact the only answers that actually make sense are either 

a) 18 (and the previous 3 expressions are erroneous); or
b) The answer is anything you like

In fact this is can be used as a good illustration of the fallacies of such "IQ" tests, and why assuming there is a fixed "answer" is not very smart in the first place.

Let me explain.

So clearly the equations are wrong from the standpoint of recognising the symbols in them as being the ones we are familiar with - that is (a) the standard decimal counting system (which we'll label SC) and the standard arithmetic operators (we'll label SO). If we assume the equations are correct, then we need to either assume SO or SC is not being used - or both. (Perhaps the symbols are from an alien world with no connection to our arithmetic system?)

First lets explore a breakdown of SO. We'll just imagine that the plus sign is the only operator being redefined, and the equals sign still means the same thing we are used to. So the plus sign is actually representing some kind of new functional operator. We can switch this out for a more digestible representation, and our equations become:

f(4) = 8
f(5) = 15
f(6) = 24
f(9) = ?

(Note we can declare the function as f(x) rather than f(x,y) since all of the statements we are considering have the same digit on either side of the + sign.)

The problem is, we can basically choose f to be absolutely anything that produces the above 3 known outputs. Perhaps the most obvious example is

(eqn 1) f(x) = x2 - 2x

and my guess is this is what the author of the question intended. Then our "answer" would be 9^2 - 2 x 9

= 63

But this definitely isn't the only function that will work. Consider e.g. a cubic, of the form

(eqn 2) f(x) = Ax3 + Bx2 + C

We then have 3 simultaneous equations

64A + 16B + C = 4
125A + 25B + C = 15
216A + 36B + C  = 24

which solve to the following values:

A = -10/37
B = 113/37
C = -1020/37

(Go ahead and try them in eqn 2). Then our answer would be

= 843/37

(or 22 and 29/37)

- maybe a bit uglier, but is there any reason this answer is any less valid? I say there isn't.

But why stop there? Consider this one, which I will dub the "anything you like polynomial":

(eqn 3) f(x) = (1/60) * [ (A - 63) x3 + (1005 - 15A) x2 + (74A - 4782) x + 7560 - 120A ]

where A is... your desired answer! Try it - choose anything you like for A, and calculate answers for f(4), f(5), f(6) and f(9) - you'll always get 8, 15, 24 - and whatever you chose for A as the answer to the last problem.

The following code illustrates this (yes I wrote it in Perl - deal with it!)


use strict;
use warnings;

my $desired_answer = 17;

my @tests = (4,5,6,9);

for my $test (@tests){
    my $result = calc( $test );
    print "x: $test, y: $resultn";

sub calc{
    my ($x) = (@_);

    my $A = $desired_answer;

    my $result = (($A - 63) * $x**3
        + (1005 - 15 * $A) * $x**2
        + (74 * $A - 4782 ) * $x
        + 7560 - 120 * $A) / 60;

    return $result;

giving the following output in the above case:

x: 4, y: 8
x: 5, y: 15
x: 6, y: 24
x: 9, y: 17

Try the code yourself (but vary $desired_answer) - you'll see how it always gives the same result for the first 3 expressions, but whatever you chose for $desired_answer as the last one. (Question for the reader: how did I find equation 3?)

Is this solution any less valid than the previous 2. I say it isn't! We are given no information about how to interpret those expressions, so how can we assume any one of the above possible solutions is the correct one?

And that's just polynomials! We didn't look at thousands of other types of function (sinusoids, gaussians, hyperbolics, ...)

And we're STILL not done! We still haven't looked at the possibility that it was the SC that was redefined instead of the SO. So here we are saying the plus sign and equals sign are good, but the digits have been redefined to mean something else.

Here's a simple example of some new algebraic rules which produce correct equations in each of the 3 expressions:

Rule 1:

The digit 1 maps to the number 2
The digit 2 maps to the number 3
The digit 5 maps to the number 10
The digits 4,6,8 map to 4,6,8 respectively (ie no change)

Rule 2:

When 2 digits are placed next to each other, it means they are multiplied
(ie 46 maps to 4 x 6 = 24)

With these 2 rules we have

4 + 4 = 8 (no change, but this equation is mathematically correct anyway)
10 + 10 = 2 * 10
6 + 6 = 3 * 4

So what about the last equation? Well if 9 maps to 9 then we need 2 numbers which multiply together to give 18 (taking into account the mapping). These work:



(because remember 1 actually means 2). Or alternatively



(remembering that 2 maps to 3). 

But of course we have no obligation to have 9 map to 9, again we can choose whatever we want. And that's really the point - if we accept the redefinition of one symbol, then we must accept that potentially all of the symbols have been redefined. Then there are quite literally infinite possibilities!

Conclusion: Maybe the people making these kinds of tests aren't sufficiently qualified to be trying to assess other people's intelligence...?!