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

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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
 
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>
+
<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.
 
which can be checked fairly easily.
  

Revision as of 13:23, 8 July 2015

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

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

Solutions

Solution 1

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

Solution 2

Let $p = (a, b, c)$, $pq = (a, b)$, $pr = (b, c)$, and $ps = (c, a)$. Then it follows that $q, r, s$ are pairwise coprime and $a = pqsa'$, $b = pqrb'$, and $c = prsc'$, with $a', b', c'$ pairwise coprime as well. Then, we wish to show \[\frac{(pqrsa'b'c')^2}{(pqrsa'b')(pqrsb'c')(pqrsc'a')} = \frac{p^2}{(pq)(pr)(ps)},\] 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

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