Question

if 1/8! + 1/9! = n/10! , find n. a)90 b)100 c)110 d)80​

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:\dfrac{1}{8!} + \dfrac{1}{9!} = \dfrac{n}{10!}$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:Value \: of \: n$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\dfrac{1}{8!} + \dfrac{1}{9!} = \dfrac{n}{10!}$

$\rm :\longmapsto\:n \: = \: \dfrac{10!}{8!} + \dfrac{10!}{9!}$

$\rm :\longmapsto\:n \: = \: \dfrac{10 \times 9 \times 8!}{8!} + \dfrac{10 \times 9!}{9!}$

$\rm :\longmapsto\:n = 90 + 10$

$\bf\implies \:n = 100$

Option (b) is correct

Note :-

Short Cut Trick :-

If x and y are natural numbers, such that

$\rm :\longmapsto\:\dfrac{x}{(y + 2)!} = \dfrac{1}{(y + 1)!} + \dfrac{1}{y!} \: then \: x= {(y + 2)}^{2}$

Here,

$\rm :\longmapsto\:\dfrac{1}{8!} + \dfrac{1}{9!} = \dfrac{n}{10!}$

$\rm :\longmapsto\:\dfrac{1}{8!} + \dfrac{1}{(8 + 1)!} = \dfrac{n}{(8 + 2)!}$

$\bf\implies \:n = {(8 + 2)}^{2} = {10}^{2} = 100$

Remark :-

$\rm :\longmapsto\:0! = 1$