What is the sum of the two smallest natural numbers, each of the two having exactly seven factors?

Correct option is D

Given:

Sum of the two smallest natural numbers, each of the two having exactly seven factors.​

Concept Used:
If a number N can be expressed as:

N = $$p_1^{a_1} \times p_2^{a_2} \times ... \times p_n^{a_n}$$​​

where $$p_1, p_2, ..., p_n$$​ are distinct prime numbers and $$a_1, a_2, ..., a_n$$​ are positive integers, then the number of factors of N is given by:

Number of factors =$$(a_1 + 1) \times (a_2 + 1) \times ... \times (a_n + 1)$$

Solution:

We want the number of factors to be 7. Since 7 is a prime number, the only way to express it as a product of integers greater than 1 is 7 itself. Therefore, we must have:

​$$a_1 + 1 = 7$$​​

​$$a_1 = 6$$​​

This means the number N must be of the form $$p^6$$​, where p is a prime number.

The two smallest prime numbers are 2 and 3. So, the two smallest numbers with exactly seven factors are:

​$$2^6 = 64$$​​

​$$3^6 = 729$$​​

The sum of these two numbers is:

64 + 729 = 793

Therefore, the sum of the two smallest natural numbers, each having exactly seven factors, is 793.