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

The correct option is A Zero

Let the numbers be $16a$ and $16b$, where $a$ and $b$ are co-primes.

We know that LCM$\times$HCF=Product of two numbers, therefore, we have,

$16a\times 16b=(16\times 146)\Rightarrow 256ab=16\times 146\Rightarrow ab=\frac{16\times 146}{256}\Rightarrow ab=9.125$

The co-primes with product $9.125$ does not exist.

Hence, the number of possible pairs are zero.