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# Using the fact that HCF (105,120) = 15, find the LCM (105,120).[a] 210[b] 420[c] 840[d] 1680. Verified
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Hint: Use the fact that if l = LCM (a,b) and g = HCF (a,b) then we have ab = lg i.e. $\text{HCF}\times \text{LCM=Product of two numbers}$.

First before using the formula we argue why the above formula is correct.
We know that if l = LCM (a,b) then any number m such that $a|m$ and $b|m\Rightarrow l\le m$.
Now we know that $g|a$ and $g|b$ so $\dfrac{a}{g},\dfrac{b}{g}\in \mathbb{N}$.
Take $m=\dfrac{ab}{g}$
$\dfrac{m}{a}=\dfrac{b}{g}\in \mathbb{N}$ i.e. $a|m,\dfrac{m}{b}=\dfrac{a}{g}\in \mathbb{N}$ i.e. $b|m$
Hence, we have $l\le m$ i.e. $l\le \dfrac{ab}{g}\Rightarrow gl\le ab\text{ (i)}$
Also we know that if g = GCD (a,b) , then any number n such that $n|a$ and $n|b\Rightarrow g\ge n$.
Take $n=\dfrac{ab}{l}$.
Since $a|l$ and $b|l$ we have $\dfrac{l}{a},\dfrac{l}{b}\in \mathbb{N}$.
Now $\dfrac{a}{n}=\dfrac{a}{\dfrac{ab}{l}}=\dfrac{l}{b}\in \mathbb{N}$ i.e. $n|a$.
Similarly, $n|b$. Hence we have $n|a$ and $n|b$.
Hence, we have $g\ge n$ i.e. $g\ge \dfrac{ab}{l}\Rightarrow gl\ge ab\text{ (ii)}$
From equation (i) and (ii) we get
\begin{align} & gl\ge ab\text{ and }gl\le ab \\ & \Rightarrow gl=ab \\ \end{align}
Hence, we have $HCF(a,b)\times LCM(a,b)=ab$.
Put a = 105, b = 120, HCF (a,b) = 15, we get
$15\times LCM(a,b)=120\times 105$
Dividing both sides by 15, we get
\begin{align} & \Rightarrow \dfrac{15LCM(a,b)}{15}=\dfrac{120\times 105}{15} \\ & \Rightarrow LCM(a,b)=840 \\ \end{align}
Hence LCM (105,120) = 840
Option [c] is correct.
Note:
 GCD is the largest of all the common divisors. Hence if m is a common divisor of a and b then $GCD\ge m$. Mathematically we write it as “that if g = GCD (a,b) , then any number n such that $n|a$ and $n|b\Rightarrow g\ge n$.” With a little bit more effort, it can be shown that $n|g$.
 LCM is the smallest of all common multiples. Hence if m is a common multiple of a and b then $LCM\le m$. Mathematically we write it as “if l = LCM (a,b), then any number m such that $a|m$ and $b|m\Rightarrow l\le m$.” With a little bit more effort it can be shown that $l|m$.
Last updated date: 29th Sep 2023
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