Identifying numbers that have the same given HCF and LCM

Question

Two numbers have a highest common factor of $12$ and a lowest common multiple of $600$. Besides $12$ and $600$ themselves , find another pair of numbers that fulfill the above condition .

I'm not sure how to carry on from my working - $$HCF= 2^2 \times 3 = 12\;\;\;\&\;\;\; LCM= 2^3 \times 3 \times 5^2 = 600 $$

Answer 1

so your pair of numbers both have 2*2*3 as a factor. you know that one of them must have 5^2 as a factor - if both had 5 as a factor, the HCF would be 60.

let these numbers be denoted a and b - a = 2*2*3 * .., b= 2*2*3 * 5*5 * ... one number must include 2^3 - so we divide both numbers by their common factor of 12, and find that

50 = ab, and b contains 5^2, or 25 - so a(b/25) = 2 , so either a or b contains 2^3.

thus, your candidates are a = 2^3 *3 = 24 or 2^2 * 3 = 12 b = 2^2 * 3 * 5^2 = 300 , and b = 2^3*3*5^2 =600

so your pairs are 24,30 or 12,600

Answer 2

There is a general method. Write $$ \begin{cases} 12 = 2^{2} \cdot 3^{1} \cdot 5^{0}\\ 600 = 2^{3} \cdot 3^{1} \cdot 5^{2}\\ \end{cases} $$ Now exchange at pleasure the powers of a fixed prime between the two lines. In this case, you can exchange the two powers of $2$ obtaining $$ \begin{cases} 24 = 2^{3} \cdot 3^{1} \cdot 5^{0}\\ 300 = 2^{2} \cdot 3^{1} \cdot 5^{2}\\ \end{cases} $$ Exchanging the two powers of $5$ will give the same result. Can you see why this method gives you all the possibilities?

Hint

Think of the definition of gcd and lcm in terms of factorizations.

Answer 3

We want to find natural numbers $a$ and $b$ such that $gcd(a,b)=12$ and $lcm(a,b)=600$.

Let $a_p$ be the exponent of $p$ in the factorization of $a$. Analogously for $b$.

Then all pairs $a,b$ come from solutions of

$\min(a_2,b_2)=2,\qquad \max(a_2,b_2)=3$

$\min(a_3,b_3)=1,\qquad \max(a_3,b_3)=1$

$\min(a_5,b_5)=0,\qquad \max(a_5,b_5)=2$

Therefore

$(a_2,b_2) \in \{ (2,3), (3,2) \}$

$(a_3,b_3) \in \{ (1,1) \}$

$(a_5,b_5) \in \{ (0,2), (2,0) \}$

a total of $2 \cdot 1 \cdot 2 = 4$ solutions, $2$ if we ignore order, and we're left with $a=24$ and $b=300$.

Answer 4

Hint If $a,b$ are your two numbers, then they are both divisible by $12$.

Write $$a=12k \\ b=12l$$ with $k,l$ relatively prime.

Then $$gcd(k,l)=1 \\ 12kl=lcm(a,b)=600$$ Therefore $kl=50$.

The problem thus reduces to writing $50$ as the product of two relatively prime numbers $k,l$.