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

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{{USAMO box|year=1972|before=First Question|num-a=2}}
{{USAMO box|year=1972|before=First Question|num-a=2}}
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[[Category:Olympiad Number Theory Problems]]
[[Category:Olympiad Number Theory Problems]]

Revision as of 17:51, 3 July 2013


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



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


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 $\alpha + \beta + \alpha - 2\alpha = \beta$. $\blacksquare$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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