Exercise 458Number theory › Divisibility and gcdHigh schoolAdded 3 Nov 2026

The least common multiple

#lcm#gcd#product relation

Statement

Let $a,b\in\mathbb{N}^*$ . Show that $\gcd(a,b)\cdot\operatorname{lcm}(a,b)=ab$ .

Hint

Write $a=\delta a'$ , $b=\delta b'$ with $\gcd(a',b')=1$ , where $\delta=\gcd(a,b)$ .

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Solution

Setup. $a=\delta a'$ , $b=\delta b'$ , $\gcd(a',b')=1$ .

LCM. Any common multiple $M$ satisfies $M=\alpha a=\beta b$ , so $\alpha a'=\beta b'$ . Since $\gcd(a',b')=1$ , $a'\mid\beta$ , say $\beta=ka'$ . Then $M=k\delta a'b'$ . The smallest positive is $k=1$ , so $\operatorname{lcm}=\delta a'b'=ab/\delta$ .

Conclusion. $\gcd(a,b)\cdot\operatorname{lcm}(a,b)=ab$ .

Remark. In particular, if $\gcd(a,b)=1$ , then $\operatorname{lcm}(a,b)=ab$ .

The least common multiple

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