Math Olympiad | Algebra Math Olympiad Problem | Find the Value of a

[@amudangopal]

Lcm missed our a+5 though it didn't matter at the end

[@mukeshmishra665]

This problem doesn't even deserve to be a part of Olympiad

[@ManojkantSamal]

(a+1)(a+2)(a+3)(a+4)
={(a+1)(a+4)}{a+2)(a+3)}
=(a^2+5a+4)(a^2+5a+6)
Let a^2+5a=R
(R+4)(R+6)
=R^2+10R+24
As per question
R^2+11R+24=0
R^2+8R+3R+24=0
R(R+8)+3(R+8)=0
(R+8)(R+3)=0
R=-8 or -3
a^2+5a=-8
a^2+5a+8=0
a= {(-5±*(25-32)}/10....not accepted
Again
R= -3
a^2+5a+3=0
a={-5±*(25-12)}/10
a={(-5)±*13)}/10, may be
*= read as square root

[@gamingwithnoob69]

Wrong