All Things Being Equal

Suppose that $2n+1$ integers have the property that upon removing any one of them, the remaining numbers can be grouped into two sets of size $n$ with the same sum. Show that all $2n+1$ numbers are equal.

Source: 1973 William Lowell Putnam Mathematical Competition

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Let us prove it by $induction$ . The statement is easily checked to be true for $n=1,2,3$ . Let it be true for $n=m$ . Consider the set $[a, a, a,... upto\ (2m+2)\ a's, b]$ with $a\leq b$ . Let us remove the first element, and group the remaining elements in two sets with $m+1$ elements in each : $[a, a, a,... upto\ (m+1)\ a's ]$ and $[a, a, a,... upto\ m\ a's, b]$ . Then $(m+1)a=ma+b$ or $b=a$ . So the statement is true for $n=m+1$

A Former Brilliant Member - 1 year, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$