Find the numbers whose L.C.M. is 180 and they are consecutive even numbers?

Hint: We will consider two consecutives even numbers 2n and 2(n+1). We can see that the number 2n and 2(n+1) has a common component of 2. We will then use the property that the product of two numbers $a$ and $b$ is equal to the product of their LCM and HCF.

Complete step by step answer:
We have to find the numbers when we have L.C.M. of two consecutives even numbers are 180.
L.C.M. = 180………(1)
We will assume the two consecutives even numbers as $2n$ and $2(n+1)$. We can absorb that $2n$ and $2(n+1)$ has 2 as common, so their H.C.F. is 2
H.C.F. = 2………(2)
We will now use the property that the product of two numbers $a$ and $b$ is equal to the product of their LCM and HCF.
$ \Rightarrow a \times b = HCF \times LCM$
We will now put the values $a$ equal to $2n$ and $b$ equals to $2(n+1)$. And then the values of H.C.F and L.C.M. from equation 1 and 2.
$ \Rightarrow 2n \times 2(n + 1) = HCF \times LCM$

We have put the value of HCF and LCM
$ \Rightarrow 2n \times 2(n + 1) = 2 \times 180$
$ \Rightarrow 4{n^2} + 4n = 2 \times 180$
We will divide the whole equation by 4
$ \Rightarrow {n^2} + n = 90$
$ \Rightarrow {n^2} + n - 90 = 0$
We will now use mid-term split property of quadratic equation to solve for n
$ \Rightarrow {n^2} + 10n - 9n - 90 = 0$

We take n and -9 common from first two term and last two terms respectively
$ \Rightarrow n(n + 10) - 9(n + 10) = 0$
We take (n+10) common
$ \Rightarrow (n + 10)(n - 9) = 0$
$ \therefore n = - 10$ and $n = 9$
We will not accept the value -10 because it is negative. We will now put the value of n in 2n and 2(n+1) to find the number

Hence, the numbers are 18 and 20 whose LCM is 180.

Note: The most common problem students face while solving this problem is to find the H.C.F. that is 2. We get 2 by using the basic definition of HCF. We should also be careful while solving the quadratic equation.