Puzzled

It all started with an apparently innocent looking puzzle which ran as

11 x 11 = 4

22 x 22 = 16

33 x 33 = ?

I thought that the progression will be (3 + 3) x (3 + 3) = 36 which is satisfying the other two and posted it as a reply to the person who sent me. I also posted in on other groups. Several replied with 36 but two persons replied with 18. Their logic was

11 x 11 = sum of the digits of the product = 1 + 2 + 1 = 4

Likewise

22 x 22 = 4 + 8 + 4 = 16

which gives

33 x 33 = 1 + 0 + 8 + 9 = 18

Well I was adamant in agreeing to it and though it too complicated but could not find an error to the logic. So my logic of simpleness as a solution was not holding ground. I would have silently ignored the replies, putting a deaf ear to the logical replies of the solvers who came with 18, allowing my false belief in my logic of simple vs complicated to cloud the logical part of my grey cells, had it not been for one of the solvers who personally messaged me and requested me to think over the solution from his perspective. Additionally, being humble in the face of my obstinate replies, he additionally asked me that he multiplied the numbers and added it while we added and then multiplied and the latter result was twice of the former. So he wanted to find whether this is a general case, where the latter operations will always return a significantly higher value than the former. I consider myself fortunate that his insistence led me to the following where not only I could find a rationale behind the results coming like he hypothesized, considering special cases of the repeated digits only, but could also find the incompleteness in the actual puzzle statement.

What follows just onwards is my findings. Once again my acknowledgement to the person, my senior in my locality, who actually triggered me to carry out this interesting work

Case36 (think this is the apt way to label this cases)

aa x aa = (a+a) x (a+a) = 4a², for a = 1, 2, 3, …., 9

Case18 (likewise!)

aa x aa = sum of digits of (10a + a)² =  sum of digits of 121a², for a = 1, 2, 3, …., 9. So let us see how many digits it can go up to. The maximum number of digits will come from max(121a²) =  9801 i.e. it will consist of 4 digits in the extreme. The sum of the highest value of 4 digits = 9 + 9 + 9 + 9 (though this will never occur) = 36.

So it may safely be said that if

4a² > 36

or,  a > 3

then Case36 will always return a higher value than Case18.

Two special situations arise for a < 3

Case36 (1) = 4

Case18 (1) = sum of digits of 121 = 4

Case36(2) = 16

Case18(2) = sum of 484 = 16

But from a = 3 onwards the values will differ. So it can also be said that the puzzle is incomplete / will have logically multiple correct ans if only conditions of a = 1, a = 2 are specified.