Difference between revisions of "1981 USAMO Problems/Problem 5"

Revision as of 10:38, 7 July 2020

Problem

Show that for any positive real $x$ , $[nx]\ge \sum_{1}^{n}\left(\frac{[kx]}{k}\right)$

Solution

We know that $x\geq\lfloor x \rfloor$ . Also, $nx\geq\lfloor nx \rfloor$ , so $x\geq\frac{\lfloor nx \rfloor}{n}$ . Thus, each of the terms in the sum is $x\geq$ , so the total sum is $nx\geq\blacksquare$

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 ==Solution==
-{{solution}}
+We know that <math>x\geq\lfloor x \rfloor</math>. Also, <math>nx\geq\lfloor nx \rfloor</math>, so <math>x\geq\frac{\lfloor nx \rfloor}{n}</math>. Thus, each of the terms in the sum is <math>x\geq</math>, so the total sum is <math>nx\geq\blacksquare</math>
 ==See Also==
 {{USAMO box|year=1981|num-b=4|after=Last Question}}

1981 USAMO (Problems • Resources)
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Difference between revisions of "1981 USAMO Problems/Problem 5"

Revision as of 10:38, 7 July 2020

Problem

Solution

See Also