Floor equation

Algebra Level pending

Find the minimum value of x x satisfying ⌊ 4 ( ⌊ x ⌋ + 1 2 ) + ⌊ x ⌋ ⌋ = 27 \left \lfloor 4 \left( \lfloor x \rfloor + \frac12\right) + \lfloor x \rfloor \right \rfloor = 27


The answer is 5.

Aly Ahmed by Aly Ahmed , 34, Egypt

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1 solution

⌊ 4 ( ⌊ x ⌋ + 1 2 ) + ⌊ x ⌋ ⌋ = 27 \displaystyle\bigg\lfloor 4\bigg(\lfloor x \rfloor + \frac{1}{2}\bigg) + \lfloor x \rfloor \bigg\rfloor = 27

⇒ 27 ≤ 4 ( ⌊ x ⌋ + 1 2 ) + ⌊ x ⌋ < 28 \Rightarrow 27 \leq 4\bigg(\lfloor x \rfloor + \frac{1}{2}\bigg) + \lfloor x \rfloor < 28

⇒ 27 ≤ 5 ⌊ x ⌋ + 2 < 28 \Rightarrow 27 \leq 5\lfloor x \rfloor + 2 < 28

⇒ 25 ≤ 5 ⌊ x ⌋ < 26 \Rightarrow 25 \leq 5\lfloor x \rfloor < 26

⇒ 5 ≤ ⌊ x ⌋ < 5 1 5 \Rightarrow 5 \leq \lfloor x \rfloor < 5\frac{1}{5}

⇒ ⌊ x ⌋ = 5 \Rightarrow \lfloor x \rfloor = 5

⇒ 5 ≤ x < 6 \Rightarrow 5 \leq x < 6

So the solution set is [ 5 , 6 ) [5,6) . The question should ask for ⌊ x ⌋ \lfloor x \rfloor for answer to be 5 5 .

0 Helpful 0 Interesting 1 Brilliant 0 Confused

Thanks. I've updated the problem statement to reflect this.

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Brilliant Mathematics Staff - 11 months, 3 weeks ago

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