Question

HCF of $3240,3600$ and a third number is 36 and their LCM is $2^4\times 3^5\times 5^2\times 7^2$ Then the third number is

• A. $2^2\times 3^5\times 7^2$
• B. $2^2\times 5^3\times 7^2$
• C. $2^2\times 5^2\times 7^2$
• D. $2^3\times 3^5\times 7^2$

Answer 1

Answer: (A)

Given:
$3240=2^3\times 3^4\times 5$
$3600=2^4\times 3^2\times 5^2$
$HCF=2^3\times 3^2\text{and}LCM=2^4\times 3^5\times 5^2\times 7^2$
Since HCF is the product of lowest power of common factors
so third number must have $2^2\times 3^2$as its factors
Since LCM is the product of highest power of common prime factors.
So third number must have $3^{2}and{7}^2$ as its factors.
Hence third number= $2^2\times 3^5\times 7^2$

Hence, Option A is correct.