Floors and roots

For discussing Olympiad Level Algebra (and Inequality) problems
ibnesina
Posts: 2
Joined: Sat Nov 24, 2012 3:36 pm

Floors and roots

find the smallest integer values of $x$ for which $\lfloor \sqrt {x +15} \rfloor - \lfloor \sqrt x \rfloor ═0$ holds
plz solve that or tell me where i can
Last edited by *Mahi* on Wed Nov 28, 2012 10:31 pm, edited 2 times in total.
Reason: LaTeXed

*Mahi*
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Re: Floors and roots

Hi ibnesina, Welcome to BdMO forum, please read and try to follow this guide http://www.matholympiad.org.bd/forum/vi ... p?f=25&t=2 on using LaTeX for writing mathematical symbols or equations.

Use $L^AT_EX$, It makes our work a lot easier!

sm.joty
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Location: Dhaka

Re: Floors and roots

you can check one by one. but the another way is that to check only squire number. Now notethat,
$x=4$ implies
$\lfloor \sqrt {x +15} \rfloor - \lfloor \sqrt x \rfloor ═1$
because $16<19<25$ so $4 <\sqrt { 4+15} <5$
now checking one by one squire number you will see that for $x=49$ you can calculate this without floor function
$\sqrt {49 +15} - \sqrt 49 ═1$
then check $x=64$ you will get your answer.

actually according to experience the trick is that, just try to make a squire for both floor function and then the next number could be your answer. just check.
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Masum
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Joined: Tue Dec 07, 2010 1:12 pm

Re: Floors and roots

ibnesina wrote:find the smallest integer values of $x$ for which $\lfloor \sqrt {x +15} \rfloor - \lfloor \sqrt x \rfloor ═0$ holds
plz solve that or tell me where i can
for $k\ge0$, $\lfloor\sqrt x\rfloor=\lfloor \sqrt{x+k}\rfloor$ if $k<2x+1$. So if $15<2x+1$ is satisfied by $8$ for smallest such $x$.
One one thing is neutral in the universe, that is $0$.

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm
One one thing is neutral in the universe, that is $0$.