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== | ||
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+ | As all of the values in the given equation are positive, we can invert it to get an equivalent equation: | ||
+ | <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> | ||
+ | We will now prove that both sides are equal, and that they are integers. | ||
+ | |||
+ | 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>. | ||
+ | WLOG let <math>\alpha \leq \beta \leq \gamma</math>. | ||
+ | |||
+ | 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 and denote the greatest common divisor and least common multiple, respectively, of the positive integers . For example, and . Prove that
Solution
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 . Let , , and be the greatest powers of that divide , , and . WLOG let .
We can now, for each of the expressions in our equation, easily determine the largest power of that divides it. In this way we will find that the largest power of that divides the left hand side is , and the largest power of that divides the right hand side is , q.e.d.
See also
1972 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |