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We know that $$0\leq a \leq b \leq c\leq d\leq e\,\,\text{ and}\,\, a + b + c + d + e = 100$$. What would be the least possible value of $\,\,a + c + e\,\,$ ?

I apologize for poor syntax.

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up vote 8 down vote accepted

$$2(a+c+e) =a+a+c+c+e+e \geq a+b+c+d+e =100$$

With equality if and only if $a=0$, $b=c$ and $d=e$.

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Just bound the first $a$ by 0 (from below). – Erick Wong Jun 14 '12 at 18:35
would this be the least value possible? – fosho Jun 14 '12 at 18:37
i apologise for the duplicates! – fosho Jun 14 '12 at 18:49
Equivalently, $a+c+e \ge 0 + b + d$ so $a+c+e \ge \frac{100}{2} = 50$ with equality iff... – Henry Jun 14 '12 at 19:41

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