Self Referential Problem - from "What would Martin Gardner Tweet"

Source: The Math Factor (contains spoilers) - also tweeted on What Would Martin Gardner Tweet

Problem:
The number of 1′s in this paragraph is ___; the number of 2′s is ___; the number of 3′s is ____; and the number of 4′s is ___.

Bonus Research Follow-up Problem:
The number of 1′s in this paragraph is ___; the number of 2′s is ___; …..(and so on) and the number of N’s is ___.
For N=2 or 3, there are no solutions (Asking that all the numbers we fill in are between 1 and N); for N=4 there are two. For N=5 there is just one, for N=6 there are none and beyond that there is just one.
Prove it.


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Solution:

Highlight the part between the * symbols for the answer.
* For N = 4 there are exactly two fillings: (2, 3, 2, 1) and (3, 1, 3, 1). Check (2,3,2,1): the paragraph's digits are the four fixed labels 1,2,3,4 plus the blanks 2,3,2,1 - total: two 1s (label + one blank), three 2s, two 3s, one 4. Checks out. Same for (3,1,3,1).

Why only these: every blank must be 1, 2 or 3 (a blank >= 4 would force a 4-count higher than 1 and the counts spiral), and the blanks sum to 2N = 8 (each of the N digits 1..N appears once as a label, plus the N blank values count themselves... precisely: total digit occurrences = N labels + N blanks, and the blanks are exactly the counts of each digit). The first blank cannot be 1 (the digit 1 appears at least once, in its own label; if the count of 1s were 1 then no blank could be 1, but then...). A short case analysis on the first blank leaves exactly the two solutions.

General N (suyash's family, works for all N > 6): fill the first blank with N-3, the second with 3, the third with 2, the (N-3)-th blank with 2, and every other blank with 1. Verify: blanks sum to (N-3) + 3 + 2 + 2 + (N-4)*1 = 2N as required; digit 1 appears once as a label plus in the N-4 one-blanks... the count of 1s is N-3 = first blank, the count of 2s is 3 = second blank, 3s is 2, the count of (N-3) is 2, and every other digit's count is 1 - all consistent.

Solution by Taaki and suyash from the comments. *

Comments

  1. Taaki...2:21 AM, July 28, 2013

    1's is 2
    2's is 3
    3's is 2
    4's is 1

    ReplyDelete
  2. suyash8:28 AM, July 28, 2013

    2, 3, 2, 1 works. As does 3, 1, 3, 1. Basically all the blanks can have only {1,2,3}. Putting a number greater than 3 in any blank doesn't work. So, the number of 4's is 1. Also, sum of the blanks should be 8, because there are 8 numbers in all. And the first blank can not have 1. A little brute force after that.

    ReplyDelete
  3. suyash9:18 AM, July 28, 2013

    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.

    ReplyDelete
  4. Pratik Poddar1:45 AM, September 14, 2026

    Verified: with (2,3,2,1) filled in, the paragraph contains exactly two 1s, three 2s, two 3s and one 4. One of the two N=4 solutions. (replied using

    ReplyDelete
  5. Pratik Poddar1:46 AM, September 14, 2026

    Both verified - those are exactly the two N=4 solutions. And your sum constraint is the right sieve: the blanks plus the four printed digits make 2N

    ReplyDelete
  6. Pratik Poddar1:47 AM, September 14, 2026

    Correct - the unique solution family for N > 6: blanks are N-3 (for 1s), 3 (for 2s), 2 (for 3s), 1s elsewhere except a 2 for (N-3)s. The counting clo

    ReplyDelete

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