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Let f(x) = floor of \frac{2-3x}{x+5}. Find f(1)+f(2)+f(3)...+f(999)+(1000).
Suppose that |a - b| + |b - c| + |c - a| = 20. What is the maximum possible value of |a - b|?
Suppose that |a - b| + |b - c| + |c - d| + \dots + |m-n| + |n-o| + \cdots+ |x - y| + |y - z| + |z - a| = 20. What is the maximum possible value of |a - n|?
Hi;
you have to put LaTeX between
tags.
Last edited by ShivamS (2014-05-21 13:56:45)
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So it would be
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Yes. For example, his post would be:
Let f(x) = floor of
For the code, click the "Quote" button on the bottom right hand side of my post.
Last edited by ShivamS (2014-05-21 14:04:05)
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found that out... ok
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Hi Dylan;
Let f(x) = floor of \frac{2-3x}{x+5}. Find f(1)+f(2)+f(3)...+f(999)+(1000).
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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So... how are you supposed to work these type of problems?
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Hi Dylan;
Let f(x) = floor of \frac{2-3x}{x+5}. Find f(1)+f(2)+f(3)...+f(999)+(1000).
Sorry, I gave you the problem with a typo. It should be f(1000).
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Let f(x) = floor of \frac{2-3x}{x+5}. Find f(1)+f(2)+f(3)...+f(999)+f(1000)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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How are you supposed work do the other two problems?
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I am working on them using a generating function approach but no luck yet.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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