More GCD and LCM

The $\text{four}$ pairs are: $\{20,2400\};\ \{60,800\};\ \{100,480\};\ \{160,300\}.$

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To see this, consider the prime content of the two numbers $a, b$ . We have $a = 2^{a_2}\cdot 3^{a_3}\cdot 5^{a_5},\ \ \ b = 2^{b_2}\cdot 3^{b_3}\cdot 5^{b_5};$ (Other prime factors are not involved here.)

Then $\text{gcd}(a,b) = 2^{\text{min}(a_2,b_2)}\cdot 3^{\text{min}(a_3,b_3)}\cdot 5^{\text{min}(a_5,b_5)} \\ \text{lcm}(a,b) = 2^{\text{max}(a_2,b_2)}\cdot 3^{\text{max}(a_3,b_3)}\cdot 5^{\text{max}(a_5,b_5)}.$ With the given numbers, $\begin{aligned} \text{gcd}(a,b) & = 20 = 2^2\cdot 3^0 \cdot 5^1 \\ \text{lcm}(a,b) & = 2400 = 2^5\cdot 3^1 \cdot 5^2 \end{aligned}$ so that $\begin{aligned} \text{min}(a_2,b_2) & = 2 & \text{max}(a_2,b_2) & = 5 \\ \text{min}(a_3,b_3) & = 0 & \text{max}(a_3,b_3) & = 1 \\ \text{min}(a_5,b_5) & = 1 & \text{max}(a_3,b_3) & = 2 \end{aligned}$ Each of the numbers $a,b$ has either the minimum or the maximum exponent for each of its prime factors; the other number has the precise opposite. This gives four possibilities: $\begin{aligned} 2^2\cdot 3^0 \cdot 5^1 = 20 &\ \text{and}\ \ 2^5 \cdot 3^1\cdot 5^2 = 2400 \\ 2^2\cdot 3^0 \cdot 5^2 = 100 &\ \text{and}\ \ 2^5 \cdot 3^1\cdot 5^1 = 480 \\ 2^2\cdot 3^1 \cdot 5^1 = 60 &\ \text{and}\ \ 2^5 \cdot 3^0\cdot 5^2 = 800 \\ 2^2\cdot 3^1 \cdot 5^2 = 300 &\ \text{and}\ \ 2^5 \cdot 3^0\cdot 5^1 = 160 \end{aligned}$

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In general, if the GCD and LCM of two numbers are given, the answer to this kind of problem is simply $2^{\Pi\left(\frac{\text{LCM}}{\text{GCD}}\right)-1},$ where $\Pi(x)$ is the number of different prime factors in $x$ . In this case, it is $2^{\Pi(2400/20)-1} = 2^{\Pi(120)-1} = 2^{3-1} = \boxed{4}.$

Since $\gcd(a,b)=20$ , we can write $a=20a'$ and $b=20b'$ , where $\gcd(a',b')=1$ .

We have $\text{lcm}(a,b)=20a'b'=2400$ , so that $a'b'=120=2^3 \cdot 3 \cdot 5$ .

Since $\gcd(a',b')=1$ , each of the prime factors $2,3,5$ appears in exactly one of $a',b'$ . To count the unordered pairs satisfying this, we can without loss of generality say that $a'$ has fewer prime factors than $b'$ .

There is $1$ way $a'$ can have zero prime factors ( $a'=1$ ) and $3$ ways $a'$ can have one prime factor ( $a' \in \{2^3,3,5 \}$ ), for a total of $\boxed4$ unordered pairs.

Because $\gcd(a,b)=20$ we can write $a=20x$ and $b=20y$ , where x and y are coprime integers.

So, $\text{lcm}(a,b)=20 \text{ lcm}(x,y) =20xy$ , so that we know that the product $xy=120$ . $120=2^3×3×5$ has 3 different prime factors, and since x and y are coprime, all powers of each prime must be either in x, or in y. so there are $2^3$ ordered pairs, and $2^3/2=\boxed{4}$ unordered pairs.