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

(terribly non-rigorous sol... oh well)
 
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:''The following solution is non-rigorous''.
 
:''The following solution is non-rigorous''.
  
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Consider the easier question <math>\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1</math>. The solution with unique values is <math>a = 2, b = 3, c = 6</math>. If we use this format to guess for <math>a, b, c</math> in the problem, then we find that <math>a = 2 \cdot 2005, b = 3 \cdot 2005, c = 6 \cdot 2005 = 12030</math>. These fit the conditions, so the answer is <math>12030</math>.  
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Consider the easier question <math>\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1</math>. The solution with unique values is <math>a = 2, b = 3, c = 6</math>. If we use this format to guess for <math>a, b, c</math> in the problem, then we find that <math>a = 2 \cdot 2005, b = 3 \cdot 2005, c = 6 \cdot 2005 = 12030</math>. These fit the conditions, so the answer is <math>12030</math>.
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There are also another solution, if we multiply (10+4+2)/16 to <math>\frac{1}{2005}</math> we get 1/8020 + 1/3208 + 1/16040 = 1/2005
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So c = 16040
  
 
==See also==
 
==See also==
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{{PMWC box|year=2005|num-b=T1|num-a=T3}}
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{{PMWC box|year=2005|num-b=I1|num-a=I3}}

Latest revision as of 07:56, 6 August 2019

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

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
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