Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

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 $d=e$.

share|improve this answer
1  
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
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.