Blackout Math is a math puzzle in which you are given an incorrect arithmetic equation. The goal of the puzzle is to remove two of the digits and/or operators in the equation so that the resulting equation is correct. For example, given the equation 6 - 5 = 15 ^ 4/2 we can remove the digit 5 and the / operator from the right-hand side in order to obtain the correct equality 6 - 5 = 1 ^ 42. Both sides of the equation now equal to 1. Observe how removing an operator between two numbers (4 and 2) causes the digits of the numbers to be concatenated (42).
Blackout Math is a math puzzle in which you are given an incorrect arithmetic equation. The goal of the puzzle is to remove two of the digits and/or operators in the equation so that the resulting equation is correct. For example, given the equation 6 - 5 = 15 ^ 4/2 we can remove the digit 5 and the / operator from the right-hand side in order to obtain the correct equality 6 - 5 = 1 ^ 42. Both sides of the equation now equal to 1. Observe how removing an operator between two numbers (4 and 2) causes the digits of the numbers to be concatenated (42). Here is a more complicated example: 288 / 24 x 6 = 18 x 13 x 8 We can remove digits and operators from either side of the equals sign (either both from one side, or one on each side). In this case, we can remove the 2 from the number 24 on the left-hand side and the 1 from the number 13 on the right-hand side to obtain the correct equality 288 / 4 x 6 = 18 x 3 x 8 Both sides of the equation now equal to 432. Here is another puzzle for you to try on your own: 168 / 24 + 8 = 11 x 3 - 16 In this question, we will write Python code to automatically solve a Blackout Math puzzle, and also to create entirely new Blackout Math puzzles.
do_op_for_numbers(op, num1, num2): Takes a string corresponding to a mathematical operation (either '+', '-', 'x', '/' or '^', and two integers, and returns the result of the operation applied to the two numbers as an integer. If the operation is division, the result should be rounded down to the nearest integer. If the operation is division and the result is infinity, return 0 instead.
>>> do_op_for_numbers('+', 3, 5)
8
>>> do_op_for_numbers('x', 4, 10)
40
>>> do_op_for_numbers('-', 1, 2)
-1
create the function : do_op_for_numbers(op, num1, num2)
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