Question

If L.C.M and H.C.F of 2 numbers are 108 and 9 respectively and one of them is 36; then find the other number.

Answer 1

Hint: For solving the question, we should know the relation between the L.C.M, H.C.F of the two numbers and the two numbers. It is stated that the product of L.C.M and H.C.F is equal to the product of the two numbers. This property is valid only for two numbers. Numerically
Let L.C.M = L and H.C.F = H of two numbers a, b. Then
L $\times $ H = a$\times $b$\to (1)$
By applying this property in the above question, we get the required number.

Complete step-by-step answer:
Let the numbers be a, b. Let the L.C.M and H.C.F of the numbers are l and h respectively.
From the property of H.C.F which states that the quotients that we get after dividing the numbers with their H.C.F are coprime numbers, we can write that
$\begin{align}
  & \dfrac{a}{h}=x\text{ ; }\dfrac{b}{h}=y\text{ where x and y are coprime numbers} \\
 & a=xh\text{ ; b=yh} \\
\end{align}$
Finding the L.C.M using ladder method
\[\begin{align}
  & x\left| \!{\underline {\,
  xh,\text{yh} \,}} \right. \\
 & y\left| \!{\underline {\,
  h,\text{yh} \,}} \right. \\
 & h\left| \!{\underline {\,
  h,\text{h} \,}} \right. \\
 & \text{ }\left| \!{\underline {\,
  1,\text{1} \,}} \right. \\
\end{align}\]
L.C.M = $(x)\times (y)\times (h)$
$\begin{align}
  & L.C.M\times H.C.F=(x)\times (y)\times (h)\times (h) \\
 & =\left( xh \right)\times \left( yh \right) \\
\end{align}$
= a * b
=Product of the two numbers.
$\therefore $L.C.M * H.C.F = a * b.
In the given question
L.C.M = 108
H.C.F = 9
a = 36
We have to find b.
Applying the formula
L.C.M * H.C.F = a * b.
108$\times $9 = 36$\times $b
Dividing by 36 gives
$\begin{align}
  & \dfrac{108\times 9}{36}=b \\
 & b=\dfrac{108}{4} \\
 & b=27 \\
\end{align}$
So, the required number is 27.

Note: We can verify the answer by calculating the L.C.M and H.C.F of the numbers 36 and 27.
L.C.M is
\[\begin{align}
  & 9\left| \!{\underline {\,
  \text{36,27} \,}} \right. \\
 & 4\left| \!{\underline {\,
  \text{4,3} \,}} \right. \\
 & 3\left| \!{\underline {\,
  \text{1,3} \,}} \right. \\
 & \text{ }\left| \!{\underline {\,
  1,\text{1} \,}} \right. \\
\end{align}\]
L.C.M = $9\times 4\times 3$ = 108
H.C.F is
\[\begin{align}
  & 9\left| \!{\underline {\,
  \text{36,27} \,}} \right. \\
 & \text{ }\left| \!{\underline {\,
  \text{4,3} \,}} \right. \\
\end{align}\]
H.C.F = 9
The L.C.M and H.C.F of 36 and 27 are 108 and 9 respectively.
Hence verified.