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April 9th, 2018, 08:19 AM   #1
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System with gcd and lcm

Can you help to solve this?
$\displaystyle \begin{cases} a+b=120 \\ \langle a,b\rangle =6(a,b) \end{cases}$
Where $\displaystyle \langle a,b\rangle$ is lcm
(a,b) gcd respectively.
a,b are natural
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April 9th, 2018, 08:57 AM   #2
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Math Focus: Algebraic Number Theory, Arithmetic Geometry
This isn't too hard if you recall $ab = \langle a,b \rangle (a,b)$. One way to solve it:

Using the equality I just stated, $\langle a,b\rangle =6(a,b)$ is equivalent to $ab = 6(a,b)^2$. Now if we write $a = a' (a,b)$ and $b = b' (a,b)$, we get $a'b' (a,b)^2 = 6(a,b)^2$ so $a'b' = 6$.

Case 1: $a' = 6, b' = 1$. In this case we have $a = 6(a,b)$ and $b = (a,b)$, so $a + b = 120$ implies $6(a,b) + (a,b) = 120$, i.e. $(a,b) = \frac{120}{7}$. This is impossible, so no solutions in this case.

The case $b' = 6, a' = 1$ is dealt with the same way, with no solutions.

Case 2: $a' = 2, b' = 3$. In this case, $a = 2(a,b)$ and $b = 3(a,b)$, so $a + b = 120$ becomes $2(a,b) + 3(a,b) = 120$, i.e. $(a,b) = \frac{120}{5} = 24$. This implies $a = 2 \times 24 = 48$, $b = 3 \times 24 = 72$. We check that this is indeed a solution: $a + b = 48 + 72 = 120$, and $ab = 48 \times 72 = 6 \times 24^2 = 6 \times (a,b)^2$ both hold.

Similarly if $a' = 3, b' = 2$, we get the solution $a = 72, b = 48$.
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April 9th, 2018, 09:00 AM   #3
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Just start with gcd(a,b) ▪ lcm(a,b) = ab
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