Question

Find the HCF and LCM of the pairs of integers and verify that $LCM(a,b)\times HCF(a,b)=a\times b$ for $16$ and $80$

Answer 1

The given integers $16$ and $80$ can be factorised as follows:

$16=2\times 2\times 2\times 280=2\times 2\times 2\times 2\times 5$

$2$ and $5$ are prime numbers as these are the numbers that are only divisible by themselves.

We know that $HCF$ is the highest common factor, therefore, the $HCF$ of $16$ and $80$ is:

$HCF=2\times 2\times 2\times 2=16$

Also, the $LCM$ is the least common multiple, therefore, the $LCM$ of $16$ and $80$ is:

$LCM=2\times 2\times 2\times 2\times 5=80$

Therefore, the $HCF$ is $16$ and $LCM$ is $80$.

Now, let $a=16$ and $b=80$, then the product of $a$ and $b$ is:

$a\times b=16\times 80=\mathrm{1280.......}\left(1\right)$

Also, the product of the $LCM$ and $HCF$ of $a$ and $b$ is:

$LCM(a,b)\times HCF(a,b)=80\times 16=\mathrm{1280.......}\left(2\right)$

Hence, from equations 1 and 2, we get $LCM(a,b)\times HCF(a,b)=a\times b$.