The sum of the first two natural numbers, each having $15$ factors (including $1$ and the number itself), is

When we want to find how many factors a number has, we use its prime factorization. For a number $N = a^p \times b^q \times c^r \times ...$ where $a, b, c$ are distinct primes, the total number of factors is:

Number of factors = $(p+1)(q+1)(r+1)...$

Each factor of $N$ is formed by choosing powers of each prime from 0 up to its maximum power. For prime $a^p$, we can choose $a^0, a^1, a^2, ..., a^p$ (that's $p+1$ choices).

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Since we need exactly 15 factors, we need:

$(p+1)(q+1)(r+1)... = 15$

All ways to write 15 as a product of positive integers:

$15 = 15 \times 1$

$15 = 5 \times 3$

This gives us two possible forms:

Form 1: $(p+1) = 15$, so $p = 14$

Number looks like: $a^{14}$

Form 2: $(p+1)(q+1) = 5 \times 3$, so $p = 4, q = 2$ (or vice versa)

Number looks like: $a^4 \times b^2$ or $a^2 \times b^4$

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For Form 1: $a^{14}$

The smallest prime is 2, so: $2^{14} = 16384$

For Form 2: $a^2 \times b^4$ and $a^4 \times b^2$

To minimize the value, we assign higher powers to smaller primes:

For $a^4 \times b^2$: Use $a = 2, b = 3$

$2^4 \times 3^2 = 16 \times 9 = 144$

For $a^2 \times b^4$: Use $a = 3, b = 2$

$3^2 \times 2^4 = 9 \times 16 = 144$

These give the same result. Let's find the next smallest:

For $a^2 \times b^4$: Use $a = 2, b = 3$

$2^2 \times 3^4 = 4 \times 81 = 324$

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Comparing all forms:

$a^{14}$: Smallest is $2^{14} = 16384$

$a^4 \times b^2$: Smallest is $2^4 \times 3^2 = 144$

$a^2 \times b^4$: Next smallest is $2^2 \times 3^4 = 324$

The first two natural numbers with exactly 15 factors are 144 and 324.

Therefore, the sum = $144 + 324 = 468$