Question

the sum of two no is 84 and their hcf is 12 the no of total pair of such no will be​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

• First number be x

• Second number be y

It is given that, HCF of x and y is 12.

So, it means there exist positive integers a and b such that

$\rm\implies \:x = 12a$

and

$\rm\implies \:y = 12b$

Further given that, sum of two numbers is 84

$\rm \: x + y = 84$

$\rm \: 12a + 12b = 84$

$\rm \: 12(a + b) = 84$

$\rm\implies \:a + b = 7$

Hence,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf a & \bf b \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 1 & \sf 6 \\ \\ \sf 2 & \sf 5 \\ \\ \sf 3 & \sf 4 \end{array}} \\ \end{gathered}$

So, possible values of x and y, corresponding to values of a and b, are

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 12 & \sf 72 \\ \\ \sf 24 & \sf 60 \\ \\ \sf 36 & \sf 48 \end{array}} \\ \end{gathered}$

So, there are 3 pairs whom sum is 84 and HCF is 12.

$\rule{190pt}{2pt}$

Additional information :-

If x and y are two positive integers having HCF H and LCM L, then

• 1. H × L = x × y

• 2. L is divisible by x, y and H

• 3. H divides x, y and L

• 4. HCF of two co-prime numbers is always 1.