24. C4 (BUL) Let \( T \) be the set of ordered triples \((x, y, z)\), where \( x, y, z \) are integers with \( 0 \leq x, y, z \leq 9 \). Players \( A \) and \( B \) play the following guessing game: Player \( A \) chooses a triple \((x, y, z)\) in \( T \), and Player \( B \) has to discover \( A \)'s triple in as few moves as possible. A move consists of the following: \( B \) gives \( A \) a triple \((a, b, c)\) in \( T \), and \( A \) replies by giving \( B \) the number \( |x+y-a-b| + |y+z-b-c| + |z+x-c-a| \). Find the minimum number of moves that \( B \) needs to be sure of determining \( A \)'s triple.
24. Two moves are not sufficient. Indeed, the answer to each move is an even number between 0 and 54, so the answer takes at most 28 distinct values. Consequently, two moves give at most \(28^2 = 784\) distinct outcomes, which is less than \(10^3 = 1000\). We now show that three moves are sufficient. With the first move \((0,0,0)\), we get the reply \(2(x+y+z)\), so we now know the value of \(s = x + y + z\). Now there are several cases: (i) \(s \leq 9\). Then we ask \((9,0,0)\) as the second move and get \((9-x-y) + (9-x-z) + (y+z) = 18 - 2x\), so we come to know \(x\). Asking \((0,9,0)\) we obtain \(y\), which is enough, since \(z = s - x - y\). (ii) \(10 \leq s \leq 17\). In this case the second move is \((9, s-9, 0)\). The answer is \(z + (9-x) + |x+z-9| = 2k\), where \(k = z\) if \(x+z \geq 9\), or \(k = 9-x\) if \(x+z < 9\). In this case, the third move is \((9, s-k-9, k)\). The answer is \(|s-k-x-y| + |s-9-y-z| + |k+9-z-x| = (k-z) + (9-x) + (9-x+k-z) = 18 + 2k - 2(x+z)\), from which we find out again whether \(k\) is \(z\) or \(9-x\). Now we are easily done. (iii) \(18 \leq s \leq 27\). Then as in the first case, asking \((0,9,9)\) and \((9,0,9)\) we obtain \(x\) and \(y\).