Difference between revisions of "2005 PMWC Problems/Problem I2"
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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>. | + | 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== | ||
{{PMWC box|year=2005|num-b=I1|num-a=I3}} | {{PMWC box|year=2005|num-b=I1|num-a=I3}} |
Latest revision as of 07:56, 6 August 2019
Problem
Let , where and are different four-digit positive integers (natural numbers) and is a five-digit positive integer (natural number). What is the number ?
Solution
- The following solution is non-rigorous.
Consider the easier question . The solution with unique values is . If we use this format to guess for in the problem, then we find that . These fit the conditions, so the answer is .
There are also another solution, if we multiply (10+4+2)/16 to 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 |