Question

Let $U$ be set with number of elements in it is $2009$ and $A,B$ are subsets of $U$ with $n(A\cup B)=280$. If $n(A^{\prime }\cap B^{\prime })=x_1^3+x_2^3=y_1^3+y_2^3$ for some positive integers $x_1,x_2,y_1,y_2$ then find the value of $(x_1+x_2+y_1+y_2)$.

Answer 1

According to the problem,

$n\left(U\right)=2009$.....(1) and $n(A\cup B)=280$.......(2).

Now

$n(A^{\prime }\cap B^{\prime })$

$=n\left\{\right(A^{\prime }\cap B^{\prime }\left)\right\}$

$=n\left(\right\{(A\cup B{)}^{\prime }\}$ [Using De Morgan's law]

$=n\left(U\right)-n(A\cup B)$

$=2009-280$

$=1729$.

Now $1729$ can be expressed as ${10}^3+9^3$ and ${12}^3+1^3$.(For this particular property of $1729$, it is called Ramanujan's Number.)

Comparing with the given problem we have

$x_1=10,x_2=9$ and $y_1=12,y_2=1$.

$\therefore (x_1+x_2+y_1+y_2)=(10+9+12+1)=32$.