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2005 PMWC Problems/Problem T3

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Problem

Replace the letters a, b, c and d in the following expression with the numbers 1, 2, 3 and 4, without repetition: a+\cfrac{1}{b+\cfrac{1}{c+\cfrac{1}{d}}} Find the difference between the maximum value and the minimum value of the expression.

Solution

By the greedy algorithm, the maximum value will occur when a = 4. To maximize the fraction, we need to minimize the quantity b + \frac{1}{c + \frac{1}{d}}, so we need to minimize b; b = 1. To minimize the remnants, we want to maximize c + \frac{1}{d}, so we want to maximize c = 3; this leaves d = 2.

Following the same pattern of alternating maximums and minimums, the minimum value of the expression occurs when a = 1, b = 4, c = 2, d = 3.

Cleaning up the complex fraction, we get

a + \cfrac{1}{b + \cfrac{d}{cd+1}} = a + \cfrac{cd + 1}{bcd + b + d}

Substituting our values gives us that the maximum is \frac{43}{9}, while the minimum is \frac{38}{31}. Subtracting, our answer is \frac{991}{279}.

See also

2005 PMWC (Problems)
Preceded by
Problem T2
Followed by
Problem T4
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10
Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
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