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Prove, for any natural number $n$, that it is possible to select $2^n$ numbers from any collection of $2^{n+1}$ natural numbers such that that sum of the $2^n$ numbers is divisible by $2^n$.

I am not sure how to even begin on this

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  • $\begingroup$ It's induction. Start with n = 1 So from any group of four numbers you can select 2 that add to a number that is a multiple of 2, i.e. the sum of two numbers is even. Well, if you can pick two odd numbers the sum is even. So let's suppose you can't. That means you can't pick two odds. That means there is at most one odd. That means there are at least three evens. You can pick two evens and the sum will be even. So for n = 1 this statement is true. Now, do the induction step. $\endgroup$ – fleablood Oct 24 '15 at 23:46
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HINT: Proceed by induction on $n$. For the induction step you’ll assume that if $A$ is a set of $2^{n+1}$ integers, there is a $B\subseteq A$ such that $|B|=2^n$, and $\sum B$ is divisible by $2^n$. Then let $A$ be a set of $2^{n+2}$ integers and try to show that there is a $B\subseteq A$ such that $|B|=2^{n+1}$, and $\sum B$ is divisible by $2^{n+1}$. To do this, consider two cases.

  • If $A$ contains at least $2^{n+1}$ even numbers, let $A_0$ be a set of $2^{n+1}$ even elements of $A$, and apply the induction hypothesis to $\left\{\frac{a}2:a\in A_0\right\}$.

  • Otherwise, let $A_0$ be a set of $2^{n+1}$ odd elements of $A$, and apply the induction hypothesis to $\left\{\frac{a-1}2:a\in A_0\right\}$.

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