Find the maximum no of elements in Y given the following conditions?

Question

MyApproach

Since X={9,15,21,27,33...375}

Then,Y={9,15,21,27..upto the middle of the set of elements X}.

So, X has 62 elements.

Therefore,Y must have 31 elements.

Can Anyone guide me How to approach the problem?

Answer 1

Your answer of 31 is correct. $X=\{9+6n :0\leq n\leq 61\}$.Now $(9+6n_1)+(9+6n_2)=384\iff n_1+n_2=61.$ So for $0\leq n_1\leq 30$ at most one member of $\{9+6n_1,9+6(61-n_1)\}$ can belong to $Y$. So $Y$ has at most $31$ members.On the other hand if we do put one member of $\{9+6n_1,9+6(61-n_1)\}$ into $Y$ for each $n_1$ from $n_1=0$ up to $n_1=61$ then the condition on $Y$ is satisfied so $Y$ can indeed have $31$ members.