IIT JEE 1983 Maths - 'Adapted' - FIB to Single-Digit Integer Q8

$\begin{aligned} (1+ax)^n & = 1 + 8x + 24x^2 + ... + a^nx^n \\ & = {n \choose 0} + {n \choose 1}ax + {n \choose 2}a^2x^2 + ... + {n \choose n}a^nx^n \end{aligned}$

Equating coefficients, we have:

$\begin{aligned} {n \choose 1}a & = 8 \\ \implies na & = 8 & ...(1) \end{aligned}$

$\begin{aligned} {n \choose 2}a^2 & = 24 \\ \frac {n(n-1)a^2}2 & = 24 \\ \color{#3D99F6} n^2a^2 - na^2 & = 48 & \small \color{#3D99F6} na = 8 \quad ...(1) \\ \color{#3D99F6} 64-8a & = 48 \\ 8 - a & = 6 \\ \implies a & = 2 \\ (1): \quad 2n & = 8 \\ \implies n & = 4 \end{aligned}$

$\implies n + 2 \lfloor a \rfloor = 4+2 \lfloor 2 \rfloor = \boxed{8}$