Question

The LCM and HCF of two numbers are 1530 and 51. Find how many such pairs are possible.

• A. $2$
• B. $3$
• C. $4$
• D. $1$

Answer 1

Answer: (C)

$4$

Let the numbers be $51a$ and $51b$, where $a$ and $b$ are co-primes.

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

$51a\times 51b=(51\times 1530)\Rightarrow 2601ab=51\times 1530\Rightarrow ab=\frac{51\times 1530}{2601}\Rightarrow ab=30$

The co-primes with product $30$ are $(1,30),(2,15),(3,10)$ and $(5,6)$.

Therefore, the required numbers are $(51,1530),(102,765),(153,510)$ and $(255,306)$

Hence, the number of possible pairs are $4$.