Algebraic determination of consecutive numbers:

• Let the 8 consecutive integers be $x, x+1, x+2, x+3, x+4, x+5, x+6, x+7$.
• Sum equation:
  $(x) + (x+1) + \dots + (x+7) = 8x + \frac{7 \times 8}{2} = 8x + 28 = 108$.
• Solve for $x$:
  $8x = 108 - 28 = 80 \implies x = 10$.
• Find the largest number:
  $\text{Largest} = x + 7 = 10 + 7 = 17$.

Comparison by decimal representation:

• Lower bound: $\frac{3}{4} = 0.75$.
• Upper bound: $\frac{5}{6} \approx 0.833$.
• Evaluating the given fraction options:
  Option 1: $\frac{4}{5} = 0.80$, which satisfies $0.75 \lt 0.80 \lt 0.833$.
  Option 2: $\frac{3}{5} = 0.60$ ($\lt 0.75$).
  Option 3: $\frac{7}{5} = 1.40$ ($\gt 0.833$).
  Option 4: $\frac{6}{5} = 1.20$ ($\gt 0.833$).

Prime versus composite analysis among the given options:

• 31 is a prime number.
• 71 is a prime number.
• 101 is a prime number.
• 91 is a composite number because it is divisible by 7 and 13 ($91 = 7 \times 13$).
  Among the given options (31, 71, 91, 101), 91 is composite while all the others are prime numbers.