Henrymavill
Approach I followed is this:
A=2b I.e. b= A/2
Similiarly, C= A/6
Abc = A^3/12 which implies that for Abc to be an integer, it has to be divisible by 12 and none of the said options satisfy this, can someone tell me where I went wrong?
Posted from my mobile deviceHi Henrymavill, Your approach is fine but your inference is wrong.
In \(Abc = A^3/12 \) , for abc to be an integer it is not necessary for ABC to be divisible by 12, instead \(A^3\) should be divisible by 12.
Again , we cannot comment anything about the individual A, B, C values also .
To correct this, we need to convert the equation in a form that removes the denominator.
We know that \(A^3 = 8 B^3 and , A^3 =216 C^3 \)
Getting rid of the denominator, the eq becomes ,
\(ABC= A^3/12 = \frac{216 C^3 }{ 12 }\)
\(ABC = 18 C^3 \)
Thus we have to check the divisibility of 2 and 9 only.