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two numbers gcd and lcm are respectively 6 and 600. What is the possible pairs of two numbers?

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    $\begingroup$ Try to use the fact that $nm = \text{gcd}(n,m) \text{lcm}(n,m)$. $\endgroup$ Mar 22, 2021 at 15:24
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    $\begingroup$ 6 and 600 itself works! $\endgroup$ Mar 22, 2021 at 15:24

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If $(a,b)$ is the pair then $a=2^x3^y5^z$ and $b=2^u3^v5^t$ and the requirements are: $\min(x,u) = 1, \max(x,u) = 3, \min(y,v) = 1, \max(y,v) = 1, \min(z,t) = 0, \max(z,t) = 2$.

Therefore there are $2\times 1 \times 2=4$ possibilities for $x,y,z,u,v,t$ and they turn into the following pairs for $(a,b)$: $(600,6),(150,24),(24,150),(6,600)$

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  • $\begingroup$ Please strive not to add more dupe answers to dupes of FAQs. $\endgroup$ Mar 22, 2021 at 19:57
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@Jorge's answer should give you the way to exhaust your solutions. But I want to help you gain some insight as to why the formulas he gave actually work.

You can use a couple of interesting facts to get an idea. The first is the fundamental theorem of arithmetic, which allows you to decompose your integers into a unique product of prime numbers.

Secondly, the GCD is the "meet" of two numbers, and the LCM is the "join" of two numbers. Anytime you have a meet or join of $a$ and $b$ in mathematics, you have

$$meet(a,b) \times join(a,b) = a \times b$$

So:

  • we have $a \times b = 3600 = 36 \times 100 = 6^2 \times 10^2 = 2^4 \times 3^2 \times 5^2$;

  • we have $meet(a, b) = 2 \times 3$;

  • and we have $join(a, b) = 2^3 \times 3 \times 5^2$,

by the fundamental theorem of arithmetic.

For all your solutions, both $a$ and $b$ will have the factors of the meet in common, and the factors of the join-divided-by-the-meet be unique to one of the factors. Also, the sum of powers of each factors of $a$ and $b$ must sum to the exponent for the same factor in $a \times b$ (because $a^b a^c = a^{b+c}$). So for example, one such pair is $a = 2 \times 3 \times 5^2 = 150$ and $b = 2^3 \times 3 = 24$.

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Hint: Setting $a=6a',\;b=6b'$, you know that $$a'b'=\frac{ab}{36}=\frac{6\cdot 600}{36}=100$$ and $a', b'$ are coprime.

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Let $m$ and $n$ be two positive integers whose gcd and lcm are $6$ and $600$ respectively.

Then, by dividing $m$ and $n$ by $6$, one reduces the problem to finding the number of unitary divisors of $100$.

The answer is $4$, since $100$ has exactly two distinct prime factors.

The four unitary divisors of $100$ ($1, 4, 25,$ and $100$) then correspond to the $(m,n)$ pairs $(6,600), (24,150), (150,24),$ and $(600,6)$ respectively.

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