The sum of the digits of a two-digit number is 11. The difference between the number and the number obtained by interchanging the positions of its digits is 33 more than twice the difference between the digits. What is the magnitude difference between the two digits of that number?
Question
The sum of the digits of a two-digit number is 11. The difference between the number and the number obtained by interchanging the positions of its digits is 33 more than twice the difference between the digits. What is the magnitude difference between the two digits of that number?
Solution
Let's denote the tens digit as x and the units digit as y.
From the problem, we know that:
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x + y = 11 (The sum of the digits is 11)
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10x + y - (10y + x) = 33 + 2|y - x| (The difference between the number and the number obtained by interchanging the positions of its digits is 33 more than twice the difference between the digits)
Simplifying the second equation, we get:
9x - 9y = 33 + 2|y - x|
or, |x - y| = (33 + 9|y - x|) / 18
Now, we know that x + y = 11, so either x = 11 - y or y = 11 - x.
Substituting y = 11 - x into the equation |x - y| = (33 + 9|y - x|) / 18, we get:
|x - (11 - x)| = (33 + 9|(11 - 2x)|) / 18
Solving this equation will give us the values of x and y, and hence, the magnitude difference between the two digits of the number.
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