My Math Forum System with gcd and lcm

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 April 9th, 2018, 08:19 AM #1 Newbie   Joined: May 2017 From: Moscow Posts: 6 Thanks: 0 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
 April 9th, 2018, 08:57 AM #2 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 270 Thanks: 81 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 $a = 72, b = 48$. Thanks from idontknow
 April 9th, 2018, 09:00 AM #3 Senior Member   Joined: Dec 2015 From: Earth Posts: 248 Thanks: 27 Just start with gcd(a,b) ▪ lcm(a,b) = ab

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