# Find numbers from GCD and LCM [duplicate]

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

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

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 $$(a,b)$$: $$(600,6),(150,24),(24,150),(6,600)$$

• Please strive not to add more dupe answers to dupes of FAQs. Mar 22, 2021 at 19:57

@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 $$b = 2^3 \times 3 = 24$$.

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.

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.