# Using the fact that HCF (105,120) = 15, find the LCM (105,120).

[a] 210

[b] 420

[c] 840

[d] 1680.

Last updated date: 18th Mar 2023

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Answer

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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. $\text{HCF}\times \text{LCM=Product of two numbers}$.

Complete step-by-step answer:

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:

[1] 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 $n|g$.

[2] 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 $l|m$.

Complete step-by-step answer:

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:

[1] 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 $n|g$.

[2] 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 $l|m$.

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