Find the number of integers from 27000 to 106780 inclusive. 2. Find the sum of these numbers. 3. Mark statements as correct/incorrect. 4. Find the square root of given numbers.

To find the number of integers from 27000 to 106780 (inclusive), we use the formula: Last number - First number + 1.

$$\text{Number of integers} = 106780 - 27000 + 1$$ $$\text{Number of integers} = 79780 + 1$$ $$\text{Number of integers} = 79781$$

The number of integers is $\boxed{\text{79781}}$.

To find the sum of these numbers, we use the formula for the sum of an arithmetic progression: $S_n = \frac{n}{2}(a + l)$, where $n$ is the number of terms, $a$ is the first term, and $l$ is the last term.
Here, $n = 79781$, $a = 27000$, and $l = 106780$.

$$S_n = \frac{79781}{2}(27000 + 106780)$$ $$S_n = \frac{79781}{2}(133780)$$ $$S_n = 79781 \times 66890$$ $$S_n = 5336180090$$

The sum of the numbers is $\boxed{\text{5,336,180,090}}$.

Mark the statements as correct (✓) or incorrect (X):

• (i) The smallest prime number is 1. ❌ (The smallest prime number is 2.)
• (ii) The smallest even prime number is 2. ✓
• (iii) A factor of a number cannot be greater than that number. ✓
• (iv) The H.C.F. of two prime numbers is 1. ✓
• (v) The H.C.F. of two even numbers will always be an even number. ✓

The question asks for the square root (ବର୍ଗମୂଳ) of the given numbers. However, some numbers are perfect cubes rather than perfect squares. Assuming the question intends for integer roots (square or cube) where they exist:

• For 1331: This is $11^3$. The cube root is $\sqrt[3]{1331} = \boxed{11}$.
• For 6084: This is $78^2$. The square root is $\sqrt{6084} = \boxed{78}$.
• For 4913: This is $17^3$. The cube root is $\sqrt[3]{4913} = \boxed{17}$.
• For 12167: This is $23^3$. The cube root is $\sqrt[3]{12167} = \boxed{23}$.
• For 32768: This is $32^3$. The cube root is $\sqrt[3]{32768} = \boxed{32}$.
• For 8000: This is $20^3$. The cube root is $\sqrt[3]{8000} = \boxed{20}$.