Question

The difference between the LCM and HCF of the natural numbers $a\And b$ is 57. What is the minimum value of $a+b$?
(a) 22
(b) 27
(c) 31
(d) 58

Answer 1

Hint: Let us assume the two numbers $a\And b$ be $ux\And uy$ respectively where u is the HCF of the two numbers. Now, LCM of (a, b) is equal to multiplication of u and x and y so multiplication will give the answer $uxy$. After that, we are going to write the equation in which the difference between LCM and HCF is equal to 57. From this, we will find some values of “u” and then solve further to get the minimum summation of $a\And b$.

Complete step by step solution:
In the above problem, we have given two natural numbers $a\And b$ and now we are going to assume the two natural numbers in terms of HCF we get,
$\begin{align}
  & a=ux; \\
 & b=uy \\
\end{align}$
In the above, “u” is the HCF of the two natural numbers $a\And b$.
Now, we are going to write the LCM of two natural numbers as follows:
$uxy$
It is given that the difference of LCM and HCF is 57 so subtracting HCF from LCM and then equating it to 57 we get,
$uxy-u=57$
Taking u as common from the above equation we get,
$\begin{align}
  & u\left( xy-1 \right)=57 \\
 & \Rightarrow xy-1=\dfrac{57}{u}..........(1) \\
\end{align}$
The values possible for u are those values in which the R.H.S of the above equation becomes an integer. If we put 1 as the value of “u” then RHS becomes 57 and solving the above equation will give:
$\begin{align}
  & xy=57+1 \\
 & \Rightarrow xy=58 \\
\end{align}$
Two possibilities for x and y which are as follows:
$\begin{align}
  & x=2; \\
 & y=29 \\
\end{align}$
Now, putting the values of $u,x,y$ in $a\And b$ we get,
$\begin{align}
  & a=ux \\
 & \Rightarrow a=1\left( 2 \right) \\
 & \Rightarrow a=2 \\
 & b=uy \\
 & \Rightarrow b=1\left( 29 \right) \\
 & \Rightarrow b=29 \\
\end{align}$
Now, adding the two natural numbers will give:
$\begin{align}
  & a+b=29+2 \\
 & \Rightarrow a+b=31 \\
\end{align}$
The other value for “u” would be 3 so putting u as 3 in eq. (1) we get,
$\begin{align}
  & xy-1=\dfrac{57}{u} \\
 & \Rightarrow xy-1=\dfrac{57}{3} \\
 & \Rightarrow xy-1=19 \\
 & \Rightarrow xy=20 \\
\end{align}$
The possible values for x and y are as follows:
$\begin{align}
  & x=4 \\
 & y=5 \\
\end{align}$
Now, substituting the values of $u,x,y$ in the natural numbers $a\And b$ we get,
$\begin{align}
  & a=ux \\
 & \Rightarrow a=3\left( 4 \right) \\
 & \Rightarrow a=12 \\
 & b=uy \\
 & \Rightarrow b=3\left( 5 \right) \\
 & \Rightarrow b=15 \\
\end{align}$
Adding the two natural numbers we get,
$a+b=15+12=27$
We got two kinds of summations in the above solution first is 29 and the other is 27 and as we are asked to find the minimum summation so the correct answer is 27.

So, the correct answer is “Option (b)”.

Note: The mistake that could be possible is that you might stop at the first summation value only (which is 31) and you can see that one of the options also has this value so make sure you have tried other values of u also and then come to any solution.