Question

One number is 18 less than the other number. The HCF of these two numbers is 6 and the LCM is 168. What is the sum of the squares of the numbers?

Answer 1

Hint: We have to use the relation between HCF, LCM and the product of the two numbers. Once we get the product of the two numbers, we can use the expansion of the square of the difference of the two numbers to get the desired result.

Complete step-by-step answer:

We are given that one number is 18 less than the other number. Let the bigger number be x, then the smaller number would be x-18.

We will now use the formula which states that for two numbers,

HCF $\times $LCM = product of the two numbers ……..(1.1)

In this case HCF of the numbers=6

And LCM of the numbers=168

Using these values in equation (1.1) for x and (x-18), we obtain

$\begin{align}

  & 6\times 168=x\left( x-18 \right) \\

 & \Rightarrow 2x(x-18)=2\times 6\times 168=2016\text{ }........(1.3) \\

\end{align}$

Again, we need to find the sum of squares of the numbers i.e.

${{x}^{2}}+{{\left( x-18 \right)}^{2}}$

For it, we can use the expansion of square of a difference i.e. ${{\left( a-b

\right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. Taking \[a=x\]and \[b=x-18\], we obtain

$\begin{align}

  & {{\left( x-(x-18) \right)}^{2}}={{x}^{2}}+{{(x-18)}^{2}}-2x(x-18) \\

 & \Rightarrow {{(x-x+18)}^{2}}={{x}^{2}}+{{(x-18)}^{2}}-2016\text{ (from equation(1}\text{.3))} \\

 & \Rightarrow \text{1}{{\text{8}}^{2}}+2016={{x}^{2}}+{{(x-18)}^{2}} \\

 & \Rightarrow {{x}^{2}}+{{(x-18)}^{2}}=2340 \\

\end{align}$

Thus, the sum of the square of the two numbers is 2340 which is the required answer.

Note: In this case, we did not have to explicitly derive x to find the answer, we just used the HCF and LCM to find the product and used it in the expansion of the difference of the numbers. One can also try to find the value of x and solve the problem, but that would result in a lot of steps.