1982 AHSME Problems/Problem 11

Problem 11 Solution

All of the digits of the numbers must be 0 to 9. If the first and last digits are x and y, we have and or and Substituting we have and Thus and which yields 16 pairs (x, y) such that the absolute value of the difference between the x and y is However, we are not done. If 0 is the last digit ( with the pair (0, 2):) then we won't have a 4 digit number, so our real value is 15. Because our digits are distinct, there are ways to fill the middle 2 places with digits, thus by the multiplication principles (counting) there are numbers that fulfill these circumstances.