To determine how many natural numbers less than 10 divide the expression \(27^{27} - 9^{40} - 3^{79}\), we need to simplify the expression based on the prime factorization.

First, let's express each number as a power of 3:

• \(27 = 3^3\), so \(27^{27} = (3^3)^{27} = 3^{81}\).
• \(9 = 3^2\), so \(9^{40} = (3^2)^{40} = 3^{80}\).
• \(3^{79}\) remains as it is.

Substituting these back into the expression:

\(27^{27} - 9^{40} - 3^{79} = 3^{81} - 3^{80} - 3^{79}\)

This can be factored as:

\(= 3^{79}(3^2 - 3 - 1)\)

\(= 3^{79}(9 - 3 - 1)\)

\(= 3^{79} \times 5\)

Now, let's determine which natural numbers less than 10 divide \(3^{79} \times 5\). The prime factorization gives us the product of two distinct prime numbers \(3\) and \(5\):

• \(3^{79}\) indicates that the number is divisible by 3.
• \(5\) indicates divisibility by 5.

Since both prime numbers 3 and 5 divide the expression, the expression is also divisible by:

• Their product, 3 × 5 = 15, which is not less than 10, but this implies no further small prime combinations.

Four natural numbers less than 10 that divide \(3^{79} \times 5\) are:

• 1 (all numbers are divisible by 1)
• 3
• 5
• The combination of 3 and 5 is 15, include divisibility within limits, offers potential number configurations when below 10, hence individual small numbers reassert.

Thus, the correct answer is that the expression is divisible by More than 3 natural numbers less than 10.