For general N>6, the following solution works.

The number of 1?s in this paragraph is N-3; the number of 2?s is 3; the number of 3?s is 2; .. the number of N-3?s is 2. All other blanks are 1.

This solution works as long as N-3 > 3, i.e., N > 6.

Thought process

Again, the sum of the blanks has to be 2N. That implies the average in a blank is 2. This must mean most values are small.

Now, most blanks are small imply the majority of the values are 1 or 2. Most blanks cannot have 2 because then we won’t be able to satisfy all the statements ‘The numbers of K’s in the paragraph are 2’.

So, let us assume that most blanks are 1 and try possible values for the first blank. Suppose we fill the first blank with N-p. Then the (N-p)th blank will have value at least 2. A larger value will again lead to all constraints not being satisfied. So, the (N-p)th blank is 2.

Now, the second blank is at least 2. But it cannot be 2 (because then there would be three 2’s). So, the second blank is greater than equal to 3. As it turns out, it cannot be greater. So, this blank is filled with 3.

Now, the third blank can be filled with 2 and all others with 1. And the value of p can be determined by the fact that the sum is 2N. Comes out to be 3.

I agree the proof is not rigorous about why no other solution works.

]]>There is a stipulation which I missed (Colm mentions it in the podcast) that the number of cards you are working with must be less than or equal to twice the number of letters in the ice cream flavor. In my case 15 (cards) is not less than or equal to 2 x 7 (7 being the letters in “vanilla”). So I did not uncover anything unexpected. The only (trivial) comment I’d make is that if you were doing the trick live, there is a modest chance that your subject would name vanilla and then take 15 or more cards from the deck – that’s what I did. (Thanks to Colm for pointing me in the right direction.) ]]>

1’s| 5 4 5 5 4 5 5

2’s| 2 4 2 2 4 2 2

3’s| 2 1 1 2 1 1 2

4’s| 2-> 1-> 3-> 1-> 1-> 3->1-> etc.

5’s| 1 2 1 2 2 1 2

6’s| 1 1 1 1 1 1 1

7’s| 1 1 1 1 1 1 1

Perhaps a better approach would be to first prove that the procedure will always converge either in a correct solution or in an infinitely repeating loop. Then, as a second step, find the upper bound of “paragraphs” or states on that loop.

This isn’t meant to be a spoiler — I just don’t think anyone would want the Math Factor community to be working feverishly to prove an incorrect conjecture. I still find it amazing that this method (maybe) is always guaranteed to converge, sometimes even with the right answer.

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The second connection may be a bit spurious and narrower; basically what I listed in the post was a way to rewrite a purported solution to get another purported solution. Such rewriting systems are another way to look at computation– in effect all that’s happening inside your computer is a gigantic rewriting of the computers memory at every step. But there are much simpler rewriting systems that already have the full power of universal computation (This isn’t saying much more than there are simple models, period, of universal computation, since in effect, every such model more or less comes down to rewriting something or another) But this puzzle as stated *isn’t* that powerful, provably, since everything eventually gets sucked into the ABAB.. cycle or the actual solution.

For more interesting examples, check out my survey “Can’t Decide? Undecide!” at http://ams.org/notices/201003 Probably my favorite there is Mysterious Example #3.

]]>For N=7 the solution is:[spoiler]

There are 4 1’s; 3 2’s; 2 3’s; 2 4’s; 1 5’s; 1 6’s; and 1 7’s in this paragraph.

A simple algorithm will transform the solution that for any N>6 to that for N+1. Add “1 N+1’s” at the end of the paragraph after “1 N’s.” Add 1 to the count for 1’s in the solution for N. Finally, move the last count of 2 that appears in the solution for N (call it the count of K’s) to the count for the next highest integer (K+1), replacing the 2 K’s by a count of 1 K’s. Applying this rule to proceed from N=7 to N=8:

There are 5 1’s; 3 2’s; 2 3’s; 1 4’s; 2 5’s; 1 6’s; 1 7’s; and 1 8’s in this paragraph.[/spoiler]

I believe this rule will work indefinitely as N is increased.

On the podcast Chaim said that the relationship between this problem and some of Alan Turing’s work would be explained on the website. I think that fell through the cracks. I’m guessing it has something to do with self-reference as mentioned on the podcast — each individual count, which can be turned into a sentence, refers to the entire paragraph. Can someone please explain where Turning comes in? ]]>

[spoiler]

My answer is that they have the same area. Bisect the angle contained by the equal sides and you get two right triangles. For the first triangle, the hypotenuse is 5 and one side is 3, so by the Pythagorean theorem the height is 4. The other triangle has hypotenuse 5 and one side is 4, so the height is 3.

The areas of both triangles equal 12.[/spoiler] ]]>