2005 PMWC Problems/Problem I2

Problem

Let $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2005}$ , where $a$ and $b$ are different four-digit positive integers (natural numbers) and $c$ is a five-digit positive integer (natural number). What is the number $c$ ?

Solution

The following solution is non-rigorous.

Consider the easier question $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$ . The solution with unique values is $a = 2, b = 3, c = 6$ . If we use this format to guess for $a, b, c$ in the problem, then we find that $a = 2 \cdot 2005, b = 3 \cdot 2005, c = 6 \cdot 2005 = 12030$ . These fit the conditions, so the answer is $12030$ .

There are also another solution, if we multiply (10+4+2)/16 to $\frac{1}{2005}$ we get 1/8020 + 1/3208 + 1/16040 = 1/2005

So c = 16040

2005 PMWC (Problems)
Preceded by Problem I1	Followed by Problem I3
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2005 PMWC Problems/Problem I2

Problem

Solution

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