Floor Ceiling Floor

Algebra Level 4

$\left\lfloor{\sqrt{\left\lceil{\sqrt{\lfloor{\sqrt{n}}\rfloor}}\right\rceil}}\right\rfloor=3$

If all integer values of $n$ that satisfy the above equation occupy the closed interval $[a,b]$ , what is the sum $a+b$ ?

Note

$\lfloor{x}\rfloor$ is the floor function , which rounds $x$ down to the nearest integer less than or equal to $x$
$\lceil{x}\rceil$ is the ceiling function , which rounds $x$ up to the nearest integer greater than or equal to $x$

The answer is 55300.

2 solutions

Chris Lewis
Apr 23, 2020

By rephrasing the floors and ceilings as inequalities, and with a bit of careful unpicking, we get $\begin{aligned} 3 \le \sqrt{\lceil \sqrt{\lfloor \sqrt{n} \rfloor } \rceil } < 4 \\ 9 \le \lceil \sqrt{\lfloor \sqrt{n} \rfloor } \rceil < 16 \\ 8 < \sqrt{\lfloor \sqrt{n} \rfloor } \le 15 \\ 64 < \lfloor \sqrt{n} \rfloor \le 225 \\ 65 \le \sqrt{n} < 226 \\ 65^2 \le n < 226^2 \\ 65^2 \le n \le 226^2-1 \end{aligned}$

Hence $a=65^2$ and $b=226^2-1$ , and so $a+b=\boxed{55300}$ .

If $9 \leq \lceil x \rceil \leq 16$ , shouldn't $8 \leq x \leq 16$ , instead of $15$ ? Afterall, all numbers between $15$ and $16$ satisfy the condition.

Guilherme Niedu - 1 year, 1 month ago

That would be true, but the second inequality in the answer above is " $<$ ", not $\le$ ".

Chris Lewis - 1 year, 1 month ago

Yes, but it's not $\leq 15$ , it's still $<16$ . As I said any number between $15$ and $16$ satisfy the condition.

Guilherme Niedu - 1 year, 1 month ago

@Guilherme Niedu – If $\lceil x \rceil < 16$ (which I think is the line you're asking about), then we can say $\lceil x \rceil \le 15$ , because $\lceil x \rceil$ is by definition an integer.

From there we get $x \le 15$ .

Sorry if I've missed your point - hopefully this explanation was clearer!

Chris Lewis - 1 year, 1 month ago

No in question itself is given that < or = and > or = for both

dark angel gaming - 1 year ago

Tran Nguyen
Apr 28, 2020

$\lfloor\sqrt{}\lceil\sqrt{}\lfloor\sqrt{}=3$
$\sqrt{}\lceil\sqrt{}\lfloor\sqrt{}\in[3,4)$	(which numbers floor down to 3)
$\lceil\sqrt{}\lfloor\sqrt{}\in[9,16)$	(just the range above squared)
$\lceil\sqrt{}\lfloor\sqrt{}\in[9,15]$	(the result of a ceiling should be an integer)
$\sqrt{}\lfloor\sqrt{}\in(8,15]$	(8.00..1 to 15 ceiling up is in 9 to 15)
$\lfloor\sqrt{}\in(64,225]$	(just the range above squared)
$\lfloor\sqrt{}\in[65,225]$	(the result of a flooring should be an integer)
$\sqrt{}\in[65,226)$	(65 to 225.99...9 flooring down is in 65 to 225)
$n\in[4225,51076)$	(just the range above squared)
$n\in[4225,51075]$	(n is an integer)

Floor Ceiling Floor

The answer is 55300.

2 solutions

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