CTK Insights

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:

a hateful problem

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.

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