# The product of 4 distinct +ve integers a,b,c,d is 8!. The numbers also satisfy ab+a+b+1=323 and bc+b+c+1=399 then d=?

Jul 25, 2018

$7$.

#### Explanation:

We have, $a b + a + b + 1 = 323$.

$\therefore \left(a + 1\right) \left(b + 1\right) = 17 \times 19. \ldots \ldots \ldots \ldots . . \left({\ast}^{1}\right)$.

Similarly, from $b c + b + c + 1 = 399$, we get,

$\left(b + 1\right) \left(c + 1\right) = 19 \times 21. \ldots \ldots \ldots \ldots \ldots . . \left({\ast}^{2}\right)$.

Comparing $\left({\ast}^{1}\right) \mathmr{and} \left({\ast}^{2}\right)$, we can say that,

$b + 1 = 19 , a + 1 = 17 \mathmr{and} c + 1 = 21$.

$\therefore a = 16 , b = 18 , c = 20$.

$\therefore d = \frac{a b c d}{a b c}$,

=(8!)/(16xx18xx20).

$\Rightarrow d = 7$.

$\textcolor{m a \ge n t a}{\text{Enjoy Maths.!}}$