Difference between revisions of "2005 PMWC Problems/Problem I2"

(terribly non-rigorous sol... oh well)
 
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==See also==
 
==See also==
{{PMWC box|year=2005|num-b=T1|num-a=T3}}
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{{PMWC box|year=2005|num-b=I1|num-a=I3}}

Revision as of 21:10, 25 April 2014

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 $12030$.

See also

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