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For a real number $x$, $⌊x⌋$ denotes the greatest integer that is less than or equal to $x$. For example, $⌊−2.5⌋=−3$ and $⌊4⌋=4$. A positive integer $a$ has the property that: $$⌊√(a)⌋+⌊√(a+1)⌋+⌊√(a+2)⌋+⌊√(a+3)⌋+⌊√(a+4)⌋=2022$$ Determine all possible values of $a$.

What I have tried: Graphing this, and tried to use wolfram alpha, both to no avail.

Prime Mover
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TheMathematician
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1 Answers1

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Firstly we know that \begin{equation} \begin{aligned} &\sqrt{a+4}-\sqrt{a} \\ =&\sqrt{a}(\sqrt{1+\frac4a}-1) \\ \lt&\frac{2}{\sqrt{a}}. \end{aligned} \end{equation} In this question, absolutely there is $a>4$, so $\sqrt{a+4}-\sqrt{a}<1$. Then we have $\lfloor \sqrt{a} \rfloor = \lfloor \sqrt{a+1} \rfloor = \lfloor \sqrt{a+2} \rfloor = 404$ and $\lfloor \sqrt{a+3} \rfloor = \lfloor \sqrt{a+4} \rfloor = 405$. What's more, we can exactly know that $\sqrt{a+3}=405$, because $\sqrt{a+2}\lt 405\le \sqrt{a+3}$ and a is a positive integer. Therefore, $a={405}^2-3=164022$.

Of course, you can verify it programmatically:

for a in range(1, 200000):

$\qquad$sum = 0

$\qquad$for i in range(5):

$\qquad$$\qquad$sum += int((a+i)**0.5)

$\qquad$if sum == 2022:

$\qquad$$\qquad$print(a)

Federico Yu
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  • Given that this looks suspiciously like a competition problem from the current year, and most competitions don't allow eliciting outside help, it would be prudent to avoid helping until we know where the problem is from, and can confirm that it is not an ongoing competition. – Arthur Jul 15 '22 at 11:50
  • You did remind me to be careful to deal with such problems. – Federico Yu Jul 15 '22 at 13:47
  • @Arthur, this problem is from a revision booklet for a competition (not from the competition itself), but I couldn't find the answer key for the review so I ask here for the answer. – TheMathematician Jul 16 '22 at 04:56