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

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

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.



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

  2. 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.

  3. 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.


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