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Similar to Find My Number (Posted on 2023-05-15) Difficulty: 2 of 5
We are going to play a regular “find my number” game, i.e. you announce your initial guess and following my answer ( limited to: “more “, “less” or “right!”) go on to your next guess.
Usually to reveal a number from 1 to N you need no more than log2N guesses in the worst case.

How many guesses are needed if I told you that my number is a 4-digit palindrome divisible by 7 ?

Assume the worst case, of course.

See The Solution Submitted by Ady TZIDON    
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Solution solution and commentary | Comment 3 of 6 |
ct=0;
for a=10:99
  n=char(string(a));
  n=str2double([n flip(n)]);
  if mod(n,7)==0
    disp(n)
    ct=ct+1;
  end
end
ct
log(ct)/log(2)

finds 18, whose base-2 log is 4 and a fraction.

        1001
        1771
        2002
        2772
        3003
        3773
        4004
        4774
        5005
        5775
        6006
        6776
        7007
        7777
        8008
        8778
        9009
        9779
ct =
    18
ans =
          4.16992500144231
          
By the example given in the puzzle, what's sought is the number of guesses needed to determine the hidden number, rather than that necessary to actually make that number the guess. That being the case, the answer is 4 guesses.

In either game, it may actually take one more guess in order actually make the guess "official". In the original version, say, Guess a number between 1 and 4. The base 2 log of 4 is 2. Suppose the number is 4. The first two guesses are: 2 and 3 so by this time the number is known to be 4, but it's not yet guessed, so it actually takes 3 guesses to get the actual number to be a guess.

The same works here: worst case the number is 9779 and the guesses are 5005, 7007, 8008, 9009. It's now known that the number is 9779, but it takes a fifth guess to nail down the "got it".


  Posted by Charlie on 2023-05-15 08:02:01
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