Question 1210217: How many ordered pairs of positive integers (m,n) satisfy \text{lcm}[m,n] = 360 and \gcd(m,n) = 360?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Solution:
We are asked to find the number of ordered pairs of positive integers $(m,n)$ such that $\text{lcm}[m,n] = 360$ and $\gcd(m,n) = 360$.
We know that for any two positive integers $m$ and $n$, the product of their least common multiple and greatest common divisor is equal to the product of the numbers themselves:
$\text{lcm}[m,n] \cdot \gcd(m,n) = m \cdot n$
Given $\text{lcm}[m,n] = 360$ and $\gcd(m,n) = 360$, we have:
$360 \cdot 360 = m \cdot n$
$m \cdot n = 129600$
Also, since $\gcd(m,n) = 360$, it means that $m$ is a multiple of 360 and $n$ is a multiple of 360. We can write $m = 360a$ and $n = 360b$ for some positive integers $a$ and $b$ such that $\gcd(a,b) = 1$.
Substituting these into the product equation:
$(360a) \cdot (360b) = 129600$
$129600 ab = 129600$
$ab = 1$
Since $a$ and $b$ are positive integers and their product is 1, the only possibility is $a = 1$ and $b = 1$.
If $a = 1$, then $m = 360 \cdot 1 = 360$.
If $b = 1$, then $n = 360 \cdot 1 = 360$.
So the only ordered pair $(m,n)$ that satisfies the given conditions is $(360, 360)$.
Let's verify if this pair satisfies the conditions:
$\gcd(360, 360) = 360$
$\text{lcm}[360, 360] = 360$
Both conditions are satisfied. Therefore, there is only one such ordered pair.
Final Answer: The final answer is $\boxed{1}$
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