On I.Q. tests and in other places, one is often confronted with problems of the form: “What’s the next element in the following series: 1, 4, 9, …”

Technically, such questions have no right answer, because there are a multitude of ways to generate the initial elements of the series, and each way can produce a different result for how the series should continue. What is being sought is the generator for the series that is somehow the simplest, cleverest, most obvious, or most elegant. There is an esthetic at work in determining the preferred solution.

For example, for the series above, an obvious answer is 16, because the initial elements of the series are the squares of the first three natural numbers, and so the obvious way to continue the series is with the squares of the subsequent natural numbers. The series is given by s

_{n}= n

^{2}.

A similar type of question involves a mapping between a series of expressions and values. In that case, the preferred answer involves the simplest or most elegant way to interpret the expression to generate the value to which it is mapped. For example, Alberto Garcia-Luis Valencia recently posted the following problem on the LinkedIn professional social network:

“Solve if u r a genius !

11 x 11 = 4

22 x 22 = 16

33 x 33 = ???”

11 x 11 = 4

22 x 22 = 16

33 x 33 = ???”

Since there are only two initial data points, the latitude for choosing a generator for this series is extremely wide. Consider the following possible ways to express the “x” operator as a function:

1. X

2. X

3. X

4. X

_{1}(a,b) = dsum(ab)2. X

_{2}(a,b) = dsum(a)dsum(b)3. X

_{3}(a,a) = 12a/11 – 84. X

_{4}(aa,aa) = 12a – 8In X

_{1}and X_{2}, dsum(x) is the sum of the digits in the base 10 representation of the number x. I.e., dsum(11)=2. X_{3}takes advantage of the fact that in our series the numbers being operated on are always the same, and X_{4}takes advantage of the fact that in our series the numbers being operated on always have repeated digits.Note that all four versions of X generate the first two terms of the series. However, they produce different predictions for the next entry. The predictions are:

1. X

_{1}: 33 x 33 = 18 (because 33 * 33 = 1089, and the sum of these digits is 18)

2. X

_{2}: 33 x 33 = 36 (because the digits of 33 sum to 6, and 6*6 = 36)

3. X

_{3}: 33 x 33 = 28 (because 12*33/11 – 8 = 36 – 8 = 28)

4. X

_{4}: 33 x 33 = 28 (because 12*3 –8 = 36 – 8 = 28)

Which of these is to be preferred? Or, is there another solution that is preferable to all of them?

The posting on the website lead to thousands of comment replies, in which people offered different answers for the next entry. By far the most popular answer was 36, possibly indicating that X

_{2}is the generator that most people find simplest and most elegant. However, most people did not indicate how they arrived at their answer, so we can’t know what generator they had in mind. A great many people came up with other answers, including 81, 64, 256, etc. What patterns did they see in the first two entries that would lead them to these conclusions?

Personally, I preferred X

_{1}. It is shorter and simpler than X

_{2}, and does not involve any arbitrary constants. It could be argued that X

_{3}or X

_{4}is simpler than either X

_{1}or X

_{2}, as it does not involve the rather non-standard function dsum, and what could be simpler than a linear extrapolation (which is what they are)? X

_{1}and X

_{2}can be expressed in English as “the digit sum of the product” and the “product of the sums of the digits.” Expressed this way, it’s harder to see X

_{1}as necessarily simpler than X

_{2}. X

_{1}and X

_{2}also exhibit a certain degree of cleverness in noticing that summing the digits can play a role.

However, there is another way to think about this problem. We tend to see 11, 22 and 33 as two-digit decimal representations of numbers. Another way to see them is as the concatenation of one-digit numbers, and to consider that concatenation can represent the addition operator. We usually treat concatenation to mean multiplication, so this is “thinking outside the box.”

Then X

_{5}(a,b,c,d) = (a+b)*(c+d). X5(a,a,a,a) = (a+a)*(a+a) = (2a)

^{2}or 4a

^{2}.

This is mathematically equivalent to X

_{2}, but doesn’t require invoking the strange dsum function, and has the esthetically pleasing expression is English: Two a’s times two a’s (i.e., two 1’s times two 1’s = 4, two 2’s times two 2’s = 16, two 3’s times two 3’s = 36).

Thus, I will concede that this is the simpler, more elegant, and more clever solution, and I understand why so many people chose 36 as the answer. Stubbornly, though, I still like my original answer.