Question

Two numbers have \[16\] as their HCF and $146$ as their LCM. How many such pairs of numbers are there?
A) zero
B) only $1$
C) only $2$
D) many

Answer 1

Hint:The HCF of two numbers is always a factor of both the numbers. There is a relation between two numbers and their LCM and HCF. By using these two facts we can check the possibilities of such pairs of numbers.

Formula used:Product of two numbers is equal to the product of their Least common divisor (LCM) and Highest common factor (HCF)
\[x \times y = LCM(x,y) \times HCF(x,y)\] where $x$ and $y$ are any two numbers. $LCM$ is the Least Common Multiple and $HCF$ is the Highest Common Factor.

Complete step-by-step answer:
Let the two numbers be $x$ and $y$.
Given that $LCM(x,y) = 146$ and $HCF(x,y) = 16$
HCF being the highest common factor is always a factor of both the numbers.
Therefore the numbers can be written as multiples of HCF.
That is, $x = 16a,y = 16b$
Here $a$ and $b$ are co-primes since there are no any factors common to them.
Now, since the product of two numbers is equal to the product of their LCM and HCF, we have
\[x \times y = LCM(x,y) \times HCF(x,y)\]
Substituting the values for $x$, $y$, their LCM and HCF,
$ \Rightarrow 16a \times 16b = 146 \times 16$
Cancelling $16$ from both the sides we have,
$ \Rightarrow 16ab = 146$
Again dividing by $16$,
$ \Rightarrow ab = \dfrac{{146}}{{16}} = \dfrac{{73}}{8}$
But $\dfrac{{73}}{8}$ is not an integer, which is not possible since $a$ and $b$ are co-primes.
Hence this case is not possible.
So, there cannot exist two numbers with LCM $146$ and HCF $16$.

So, the correct answer is “Option A”.

Note:The LCM of two numbers must be a multiple of their HCF of those two numbers too. Here 146 is not a multiple of 16 . So we can say directly that this case is not possible.