Question

Tìm số tự nhiên a, b biết ƯCLN (a;b) = 4 và a + b = 48.

Answer 1

\(\Rightarrow\left\{{}\begin{matrix}a=4k\\b=4q\end{matrix}\right.\left(k,q\in N\text{*}\right)\\ a+b=48\Rightarrow4\left(k+q\right)=48\\ \Rightarrow k+q=12\)

Mà \(\left(k,q\right)=1\Rightarrow\left(k,q\right)\in\left\{\left(1,11\right);\left(11,1\right);\left(5,7\right);\left(7,5\right)\right\}\)

\(\Rightarrow\left(a,b\right)\in\left\{\left(4,44\right);\left(44,4\right);\left(20;28\right);\left(28;20\right)\right\}\)

Answer 2

⇒{a=4kb=4q​(k,q∈N*)a+b=48⇒4(k+q)=48⇒k+q=12

Mà $\left(k,q\right)=1\Rightarrow \left(k,q\right)\in \left\{\left(1,11\right);\left(11,1\right);\left(5,7\right);\left(7,5\right)\right\}$

$\Rightarrow \left(a,b\right)\in \left\{\left(4,44\right);\left(44,4\right);\left(20;28\right);\left(28;20\right)\right\}$