Difference between revisions of "1972 USAMO Problems/Problem 1"
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<center><math>\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}</math>.</center> | <center><math>\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}</math>.</center> | ||
− | ==Solution== | + | ==Solutions== |
+ | |||
+ | ===Solution 1=== | ||
As all of the values in the given equation are positive, we can invert it to get an equivalent equation: | As all of the values in the given equation are positive, we can invert it to get an equivalent equation: | ||
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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>. <math>\blacksquare</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>. <math>\blacksquare</math> | ||
+ | ===Solution 2=== | ||
+ | |||
+ | Let <math>p = (a, b, c)</math>, <math>pq = (a, b)</math>, <math>pr = (b, c)</math>, and <math>ps = (c, a)</math>. Then it follows that <math>q, r, s</math> are pairwise coprime and <math>a = pqsa'</math>, <math>b = pqrb'</math>, and <math>c = prsc'</math>, with <math>a', b', c'</math> pairwise coprime as well. Then, we wish to show | ||
+ | <cmath>\frac{(pqrsa'b'c')^2}{(pqrsa'b')(pqrsb'c')(pqrsc'a')} &= \frac{p^2}{(pq)(pr)(ps)},</cmath> | ||
+ | which can be checked fairly easily. | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 20:53, 31 March 2014
Problem
The symbols and denote the greatest common divisor and least common multiple, respectively, of the positive integers . For example, and . Prove that
Solutions
Solution 1
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 .
Solution 2
Let , , , and . Then it follows that are pairwise coprime and , , and , with pairwise coprime as well. Then, we wish to show
\[\frac{(pqrsa'b'c')^2}{(pqrsa'b')(pqrsb'c')(pqrsc'a')} &= \frac{p^2}{(pq)(pr)(ps)},\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)
which can be checked fairly easily.
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 (Problems • Resources) | ||
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.