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=?

Question

3738 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-25T14:04:41

#7#.

We have, #ab+a+b+1=323#.

#:. (a+1)(b+1)=17xx19...............(ast^1)#.

Similarly, from #bc+b+c+1=399#, we get,

#(b+1)(c+1)=19xx21..................(ast^2)#.

Comparing #(ast^1) and (ast^2)#, we can say that,

#b+1=19, a+1=17 and c+1=21#.

#:. a=16, b=18, c=20#.

# :. d=(abcd)/(abc)#,

#=(8!)/(16xx18xx20)#.

# rArr d=7#.

#color(magenta)("Enjoy Maths.!")#