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Given the set $M=\{2^1, 2^2, ...,2^{15}\}$ . Prove that there are no 3 disjoint sets $X, Y, Z$ such $M=X \cup Y \cup Z$ $S(X)=S(Y)=S(Z)$ where $S(W)$ is define as the sum of elements in the set $W$. I need a very, very elemetary proof, such it can be explained to a fifth grader.

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Since $2^{1} + 2^{2} + \cdots + 2^{14} = 2^{15} - 2$, whichever set contains $2^{15}$ will have a larger sum than any of the others.

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  • $\begingroup$ Proofs of $2^0+2^1+2^2+ \cdots + 2^n=2^{n+1}-1$ can be found here (binary) and here (induction) $\endgroup$
    – Air Conditioner
    Commented Feb 26, 2018 at 19:15
  • $\begingroup$ Awesome Alex . Thank you. The only proof I've seen was impling serious algebra. Thanks alot. $\endgroup$
    – motoras
    Commented Feb 26, 2018 at 19:19

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