# CTK Insights

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13 Apr

### 7 or 22

"Have you guessed the riddle yet?" the Hatter said, turning to Alice again.

"No, I give it up," Alice replied. "What's the answer?"

"I haven't the slightest idea," said the Hatter.

L. Carroll, Alice's Adventures in Wonderland

I have an instinctive dislike for the kind of questions that appear regularly on various forums and social networks. Here's an example that was posted at the CutTheKnotMath facebook page:

Without ever trying to answer such questions, I was always confident that the poster (if not the author) were smugly awaiting a definite reply, although, even with the most benevolent interpretation, the problem has to be considered ill-posed, like that of asserting the next term in a given sequence.

The question being posted at the CutTheKnotMath facebook page, I gave it some thought. The anticipated answer was likely to be

Since $(3+4=\;) 19 = 3 + 4 + 3\cdot 4$ and $(5+6=\;) 41 = 5 + 6 + 5\cdot 6,$ then $(1+3=) 1+3+1\cdot 3=7.$

Obviously, the author expected an algorithmic procedure (a formula most likely) that when applied to $3$ and $4$ led to $19$ but when applied to $5$ and $6$ produced $41.$ That algorithm had then to be applied to the pair $1,3.$

Bui Quang Tuan offered a different interpretation and wondered which is more natural:

$1 + 3 = (5 + 6)-(3 + 4) = 41-19 = 22$

That was an interesting approach to a question, most certainly overlooked by the author. (While the latter used the symbol of addition just to suggest a presence of an algorithm, Bui Quang Tuan cleverly accepted the two given identities as such and applied to them the regular operations of addition and subtraction. But even choosing a more orthodox interpretation, an algorithm that would bewilder the problem's author is not difficult to find.

Let's agree to denote the algorithm as a function of two variables. E.g., the above cold be described as $f(x,y)=x+y+x\cdot y$ with the common meaning of arithmetic operations. But here's another possibility: $g(x,y)=x\cdot y+y+y-x/x,$ such that $g(3,4)=19$ and $g(5,6)=41.$ With this interpretation $1+3=g(1,3)=1\cdot 3+3+3-1/1=8.$

Vu Xuan Hanh posted another example: $h(x,y)=x+y\cdot y.$ Indeed, $h(3,4)=3+4\cdot 4=19$ and $h(5,6)=5+6\cdot 6=41,$ implying that $1+3=h(1,3)=1+3\cdot 3=10.$

The latter example has caused a shift in my view of the problem. Perhaps, it is less like finding the next term of a sequence than expressing various integers with a fixed set of numbers using various arithmetic operations. For example, if the task is to place the symbols of arithmetic operations of parentheses between the digits $624663$ so as to get, say, $20$ as the result, we get the possibilities: $6\cdot 2+4+6+6/3=20,$ $(6-2)\cdot 4+6/6+3=20,$ $-(6-2)/(4-6)+6\cdot 3=20,$ $6-2+4+6\cdot 6/3.$ There bound to be more.

Why do I prefer the latter kind of problems to the one I considered at the beginning? The difference between them is in the veil of mystery with which one is presented. There is always an implicit stipulation that somehow the problem has a unique solution. It often comes with an enticing warning "99% of the population can't do that right. Can you?" But, obviously, it would be a rare circumstance where the problem of this kind has a unique solution. With this realization, the problem becomes less obnoxious and can be looked at from other angles. For example,

1. Find several formulas $f(x,y)$ for which $f(3,4)=19$ and $f(5,6)=41$. See what is $f(1,3).$
2. For a given formula $f(x,y)$ such that $f(3,4)=19$ and $f(5,6)=41,$ find which other integers can be represented as $f(x,y).$
3. For which integers $n$ there are $a,\;$ $b,\;$ and $f(x,y),$ such that $f(3,4)=19,$ $f(5,6)=41,$ and $f(a,b)=n.$
4. Are there integers $n$ provably not expressible as $f(x,y),$ where $f(3,4)=19,$ $f(5,6)=41,$ perhaps for a given $f.$