What is the highest common factor (HCF) of $2^4\times 3^2\times 5^3\times 13$ and $2\times 3\times 5\times 11\times {13}^2$?

The correct answer is 2×3×5×13.

To find the Highest Common Factor (HCF) of numbers expressed in their prime factorized form, we take the product of the lowest power (smallest exponent) of each common prime factor present in the numbers.

Given the two numbers:

$\text{First number}=2^4\times 3^2\times 5^3\times 13$

$\text{Second number}=2\times 3\times 5\times 11\times {13}^2$

Let us analyze each prime factor step-by-step:

1. Prime factor 2:
The exponents of 2 in the two numbers are 4 and 1. The smaller exponent is 1.
$\text{Lowest power}=2^1$

2. Prime factor 3:
The exponents of 3 in the two numbers are 2 and 1. The smaller exponent is 1.
$\text{Lowest power}=3^1$

3. Prime factor 5:
The exponents of 5 in the two numbers are 3 and 1. The smaller exponent is 1.
$\text{Lowest power}=5^1$

4. Prime factor 13:
The exponents of 13 in the two numbers are 1 and 2. The smaller exponent is 1.
$\text{Lowest power}={13}^1$

Note: The prime factor 11 is not present in the first number, so it is not a common factor and is not included in the HCF.

Multiplying these lowest powers of common prime factors together, we obtain:

$\text{HCF}=2\times 3\times 5\times 13$