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If $$x\lfloor x\rfloor =39 \quad \text{and}\quad y\lfloor y \rfloor=68.$$

What is the value of:

$$\lfloor x\rfloor+\lfloor y \rfloor $$

I don't know how to solve such problems.

I would appreciate an insight regarding the general approach to such problems.

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    $\begingroup$ observe that $[y] \leq y < [y]+1$ and $[y]$ is an integer by definition. use this observation to solve for $[y]$. do the same for x $\endgroup$ – mm-aops Jun 13 '13 at 19:24
  • $\begingroup$ These involve the floor function, not absolute value. $\endgroup$ – Ross Millikan Jun 13 '13 at 19:26
  • $\begingroup$ yeah, that's what I mean by $[x]$ I write $|x|$ for abs value, although I agree it's hard to distinguish them in this font $\endgroup$ – mm-aops Jun 13 '13 at 19:29
  • $\begingroup$ Hint: $x \lfloor x \rfloor$ is close to $x \cdot x = x^2$. $\endgroup$ – Shaun Ault Jun 13 '13 at 19:30
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    $\begingroup$ $14$ is my answer. $\endgroup$ – Adienl Jun 13 '13 at 19:31
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Notice that $\lfloor x\rfloor$ is not very different from $x$, so $x\lfloor x\rfloor $ is not much different from $ x^2$. If you want $x\lfloor x\rfloor = 39$, you need $x^2$ to be about $ 39$ also, which means $x$ is going to be around 6 or so, and $\lfloor x\rfloor$ will be exactly 6. Then $x=6\frac12$ does the trick.

Do the same for $y$, and then add the results.

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    $\begingroup$ We don't even need $x,y$ just their floors. $\endgroup$ – vadim123 Jun 13 '13 at 20:24
  • $\begingroup$ If you don't calculate $x$, you need a more complicated argument to show that the $\lfloor x\rfloor$ that you have is actually the floor of a solution. Suppose you had guessed $\lfloor x \rfloor = 7$. It's not hard to show that this is wrong. But it's harder than showing that $x=6\frac12$ is right. $\endgroup$ – MJD Jun 14 '13 at 13:12
  • $\begingroup$ All you need is $6^2\le 39<7^2$ to get $\lfloor x\rfloor=6$. No need to determine $6.5$ exactly. $\endgroup$ – vadim123 Jun 14 '13 at 15:37
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Note that, because $$ \lfloor x\rfloor \leq x $$ then, assuming positive values for x: $$ \lfloor x\rfloor ^2 \leq \lfloor x\rfloor x \leq x^2 $$ The center term is the value they gave us (39), so $\lfloor x\rfloor ^2$ must be some "perfect" or integer square less than that.

Now, there's the temptation to just grab the highest perfect square less than 39 (which would be 36), but we need to be careful; we need to consider the possibility that, in reducing $x$ to $\lfloor x\rfloor$ we could have reduced $\lfloor x\rfloor ^2$ past 36. How can we be sure that $\lfloor x\rfloor ^2$ is the highest perfect square below 39? Well, consider this...

If $\lfloor x\rfloor$ were some integer less than 6, then x would have to be less than 6, so $x^2$ would have to be less than 36, which would violate the right half of $\lfloor x\rfloor ^2 \leq \lfloor x\rfloor x \leq x^2$ (because 39 isn't less than nor equal to 36). You can do this symbolically and show that this principle holds true for any $x \geq -1$, but I didn't want to make things too complicated. Suffice it to say that $\lfloor x\rfloor ^2$ will be the highest perfect square which is less than or equal to the value given for $\lfloor x\rfloor x$.

This yields: $\lfloor x\rfloor ^2 = 36$, and $\lfloor y\rfloor ^2 = 64$, and then you're home free.

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$$ x |x| = 39 $$

Means $x$ is at the very least positive, since otherwise 39 could not be positive:

$$ -x |-x| = -xx = -39 $$

So this is pretty much equivalent, but not equal to:

$$x^2 = 39$$

Likewise for the other expression. So the result should be something like:

$$\sqrt{39} + \sqrt{68}$$

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  • $\begingroup$ I think the question seems to be about $\lfloor x \rfloor$, not $|x|$. $\endgroup$ – MJD Jun 13 '13 at 20:31

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