Let \( f(x, y) = x + y + xy. \) It can be rewritten as \( f(x, y) =
(x + 1)(y + 1) - 1. \) Now, consider an arbitrary set of integers
\( A = \{a_1, a_2, \dots, a_n\}. \) Applying the algorithm
mentioned in the problem statement to two integers \( a_i \) and \(
a_j \) chosen randomly from the set such that \( i \ne j, \) we
obtain, \( f(a_i, a_j) = (a_i + 1)(b_i + 1) - 1. \) When this
expression is used in subsequent calls to \( f(x, y), \) the \( -1
\) drops, the new number is included in the product and the \( -1 \)
gets added again. For instance, using this result and a different
number \( a_k \) chosen randomly from the set, we obtain \(
f\bigl((a_i + 1)(a_j + 1) - 1, a_k\bigr) = (a_i + 1)(a_j + 1)(a_k +
1) - 1. \) Inductively, this pattern holds, and the algorithm
returns a product of all integers in the set minus \( 1. \)
Therefore, the result of the algorithm, when applied to the set \( A
\) is \( \prod_{i=1}^n (a_i + 1) - 1. \) If \( A = \{1, 2, \dots,
9\}, \) on applying the algorithm, we obtain \( (1 + 1)(2 + 1) \dots
(9 + 1) - 1 = 10! - 1 = 3628799. \)
Ryan Batterman said:
Let \( f(x, y) = x + y + xy. \) It can be rewritten as \( f(x, y) = (x + 1)(y + 1) - 1. \) Now, consider an arbitrary set of integers \( A = \{a_1, a_2, \dots, a_n\}. \) Applying the algorithm mentioned in the problem statement to two integers \( a_i \) and \( a_j \) chosen randomly from the set such that \( i \ne j, \) we obtain, \( f(a_i, a_j) = (a_i + 1)(b_i + 1) - 1. \) When this expression is used in subsequent calls to \( f(x, y), \) the \( -1 \) drops, the new number is included in the product and the \( -1 \) gets added again. For instance, using this result and a different number \( a_k \) chosen randomly from the set, we obtain \( f\bigl((a_i + 1)(a_j + 1) - 1, a_k\bigr) = (a_i + 1)(a_j + 1)(a_k + 1) - 1. \) Inductively, this pattern holds, and the algorithm returns a product of all integers in the set minus \( 1. \)
Therefore, the result of the algorithm, when applied to the set \( A \) is \( \prod_{i=1}^n (a_i + 1) - 1. \) If \( A = \{1, 2, \dots, 9\}, \) on applying the algorithm, we obtain \( (1 + 1)(2 + 1) \dots (9 + 1) - 1 = 10! - 1 = 3628799. \)