Difference between revisions of "1972 USAMO Problems/Problem 1"

(New page: ==Problem== The symbols <math>(a,b,\ldots,g)</math> and <math>[a,b,\ldots, g]</math> denote the greatest common divisor and least common multiple, respectively, of the positive integers <m...)
 
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==Solution==
 
==Solution==
{{solution}}
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As all of the values in the given equation are positive, we can invert it to get an equivalent equation:
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<center><math>\frac{[a,b][b,c][c,a]}{[a,b,c]^2}=\frac{(a,b)(b,c)(c,a)}{(a,b,c)^2}</math>.</center>
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We will now prove that both sides are equal, and that they are integers.
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Consider an arbitrary prime <math>p</math>. Let <math>p^\alpha</math>, <math>p^\beta</math>, and <math>p^\gamma</math> be the greatest powers of <math>p</math> that divide <math>a</math>, <math>b</math>, and <math>c</math>.
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WLOG let <math>\alpha \leq \beta \leq \gamma</math>.
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We can now, for each of the expressions in our equation, easily determine the largest power of <math>p</math> that divides it. In this way we will find that the largest power of <math>p</math> that divides the left hand side is <math>\beta+\gamma+\gamma-2\gamma = \beta</math>, and the largest power of <math>p</math> that divides the right hand side is <math>\alpha + \beta + \alpha - 2\alpha = \beta</math>, q.e.d.
  
 
==See also==
 
==See also==

Revision as of 12:15, 31 January 2009

Problem

The symbols $(a,b,\ldots,g)$ and $[a,b,\ldots, g]$ denote the greatest common divisor and least common multiple, respectively, of the positive integers $a,b,\ldots, g$. For example, $(3,6,18)=3$ and $[6,15]=30$. Prove that

$\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}$.

Solution

As all of the values in the given equation are positive, we can invert it to get an equivalent equation:

$\frac{[a,b][b,c][c,a]}{[a,b,c]^2}=\frac{(a,b)(b,c)(c,a)}{(a,b,c)^2}$.

We will now prove that both sides are equal, and that they are integers.

Consider an arbitrary prime $p$. Let $p^\alpha$, $p^\beta$, and $p^\gamma$ be the greatest powers of $p$ that divide $a$, $b$, and $c$. WLOG let $\alpha \leq \beta \leq \gamma$.

We can now, for each of the expressions in our equation, easily determine the largest power of $p$ that divides it. In this way we will find that the largest power of $p$ that divides the left hand side is $\beta+\gamma+\gamma-2\gamma = \beta$, and the largest power of $p$ that divides the right hand side is $\alpha + \beta + \alpha - 2\alpha = \beta$, q.e.d.

See also

1972 USAMO (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5
All USAMO Problems and Solutions
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