Prove by induction : if n < =(2^k) -1 then we can find the number with k guesses
To prove this statement using induction, we need to show that it holds true for a base case and then prove the inductive step.
Base case: For k = 1, we have (2^k) - 1 = (2^1) - 1 = 1. In this case, if n ≤ 1, then we can find the number with 1 guess, as there is only one possible number - 1.
Inductive step: Assume that for some value of k = m, the statement holds true, i.e., if n ≤ (2^m) - 1, then we can find the number with m guesses. We will prove that it is also true for k = m + 1.
For k = m + 1, we have (2^k) - 1 = (2^(m+1)) - 1 = 2*(2^m) - 1. Let's consider the range of numbers from 1 to (2^m) - 1 as the first half, and the range from (2^m) to (2^(m+1)) - 1 as the second half.
To find a number within the range of (2^m) to (2^(m+1)) - 1, we can first make a guess for the middle number in this range, which is (2^m) + ((2^m) / 2) = (3/2)*(2^m).
If we guess a number less than or equal to (3/2)*(2^m), we know that the number lies in the first half and we can continue the process of guessing with m guesses or less, according to our assumption. This is because (3/2)*(2^m) - 1 ≤ (2^m) - 1.
If we guess a number greater than (3/2)*(2^m), we know that the number lies in the second half. However, since the second half has the same size as the first half (i.e., (2^m) - 1 numbers), we can still apply our assumption and find the number with m guesses or less.
Therefore, by the principle of mathematical induction, the statement holds for all positive integers k.
In conclusion, if n ≤ (2^k) - 1, then we can find the number with k guesses.