2005 PMWC Problems/Problem T3

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 $\frac{991}{279}$ .

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

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2005 PMWC Problems/Problem T3

Problem

Solution

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