A Pre-RMO question! -8

Given that $\begin{cases} ab+a+b = 322 & ...(1) \\ bc + b + c = 398 & ...(2) \end{cases}$ $\implies (2)-(1): \quad (b+1)(c-a) = 76 = 4 \times 19$ .

$\implies b + 1 = \begin{cases} \begin{array} {rllll} 4 & \implies b = 3 & (1): \quad 3a + a + 3 = 322 & \implies \red{a = \dfrac {319}4} & \small \red{\text{Not an integer}} \\ 19 & \implies b = 18 & (1): \quad 18a + a + 18 = 322 & \implies \blue{a = 16} & \small \blue{\text{Acceptable}} \end{array} \end{cases}$

From $(b+1)(c-a) = 4\times 19$ , since $b = 18$ , $c-a = 4 \implies c = 4 + 16 = 20$ . Also given that $abcd = 40320$ , $\implies d = \dfrac {40320}{16\times 18 \times 20} = \boxed 7$ .

We have $b=\frac{323}{a+1}-1$ .

So $323=17\times 19$ must be divisible by $a+1$ . Hence, $a=16,b=18$ or $a=18,b=16$ .

And $c=\frac{399}{b+1}-1$ .

So $399=19\times 3\times 7$ must be divisible by $b+1$ . Hence we get six possible values of $b$ , of which only $b=18$ is common with the earlier value.

So, $a=16,b=18,c=20$ and $d=\frac{40320}{16\times 18\times 20}=\boxed 7$ .

We are given $\begin{cases} ab + a + b = 322 \\ bc + b + c = 398 \end{cases} \implies \begin{cases} ab + a + b + 1 = 323 \\ bc + b + c + 1 = 399 \end{cases} \implies \begin{cases} (a + 1)(b+1) = 323 \\ (b+1)(c+1) = 399 \end{cases}$

Both equations have a common factor $(b+1)$ . And the common factors of $323$ and $399$ are only $1, 19$

$b+1 = \begin{cases} 1 \implies \color{#D61F06}{b = 0 \quad \quad [b \text{ is positive integer}]} \\ 19 \implies \color{#3D99F6}{b = 18} \end{cases}$

Now from the conditions, we can find values of $a$ and $c$ : \(\begin{cases} (a + 1)(b+1) = 323 \\ (b+1)(c+1) = 399 \end{cases}

\implies

\begin{cases} (a + 1) = \frac{323}{19} \\ (c+1) = \frac{399}{19} \end{cases}

\implies

\begin{cases} a = 17 -1 \\ c = 21 - 1 \end{cases}

\implies

\begin{cases} a = 16 \\ c = 20 \end{cases} \)

$abcd = 40320 \implies d = \dfrac{40320}{16 \times 18 \times 20} \implies \boxed{d = 7}$