Solve this floor function

Algebra Level 5

1 ⌊ x ⌋ + 1 ⌊ 2 x ⌋ = { 9 x } + 1 3 \large \frac 1{\lfloor x \rfloor} + \frac 1{\lfloor 2 x \rfloor} = \{ 9x \} + \frac 13

If the smallest positive solution of x x satisfying the equation above is of the form m n \frac {m}{n} , where m m and n n are coprime positive integers, find m − n m-n .

Notations :


The answer is 113.

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Md Zuhair by Md Zuhair , 20, India

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1 solution

1 ⌊ x ⌋ + 1 ⌊ 2 x ⌋ = { 9 x } + 1 3 1 ⌊ x ⌋ + 1 ⌊ 2 x ⌋ − 1 3 = { 9 x } \begin{aligned} \frac 1{\lfloor x \rfloor} + \frac 1{\lfloor 2 x \rfloor} & = \{ 9x \} + \frac 13 \\ \frac 1{\lfloor x \rfloor} + \frac 1{\lfloor 2 x \rfloor} - \frac 13 & = \{ 9x \} \end{aligned}

Since 0 ≤ { 9 x } < 1 0\le \{ 9x \} < 1 , for integer x x , we have:

0 ≤ 1 x + 1 2 x − 1 3 < 1 1 3 ≤ 1 x + 1 2 x < 4 3 1 3 ≤ 3 2 x < 4 3 3 4 < 2 x 3 ≤ 3 3 4 < 2 x 3 ≤ 3 2 ≤ x ≤ 4 \begin{aligned} 0 \le \frac 1x & + \frac 1{2 x} - \frac 13 < 1 \\ \frac 13 \le \frac 1x & + \frac 1{2 x} < \frac 43 \\ \frac 13 \le & \ \frac 3{2 x} < \frac 43 \\ \frac 34 < & \ \frac {2 x}3 \le 3 \\ \frac 34 < & \ \frac {2 x}3 \le 3 \\ 2 \le & \ x \le 4 \end{aligned}

Therefore, the smallest solution x > ≈ 2 x > \approx 2 and we have:

1 ⌊ x ⌋ + 1 ⌊ 2 x ⌋ − 1 3 = 1 2 + 1 4 − 1 3 = 5 12 ⟹ { 9 x } = 5 12 { 9 ( 2 + { x } ) } = 5 12 9 { x } = 5 12 ⟹ { x } = 5 108 ⟹ x = 2 + 5 108 = 221 108 \begin{aligned} \frac 1{\lfloor x \rfloor} + \frac 1{\lfloor 2 x \rfloor} - \frac 13 & = \frac 12 + \frac 14 - \frac 13 = \frac 5{12} \\ \implies \{ 9 x\} & = \frac 5{12} \\ \big \{ 9 (2 + \{ x \}) \big \} & = \frac 5{12} \\ 9 \{ x \} & = \frac 5{12} \\ \implies \{ x \} & = \frac 5{108} \\ \implies x & = 2 + \frac 5{108} = \frac {221}{108} \end{aligned}

⟹ m − n = 221 − 108 = 113 \implies m - n = 221-108 = \boxed{113}

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@Chew-Seong Cheong Please note that you are not obliged to spoonfeed everyone that harasses you (even though I am appreciative of your efforts in doing so). It should be on them to demonstrate what they have tried, instead of demanding people give them the complete answer.

@Md Zuhair Please refrain from doing this in future, as it is very disruptive to the community.

Calvin Lin Staff - 4 years, 8 months ago

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Okay sir, I am very sorry, I will not do it from now on,

Md Zuhair - 4 years, 2 months ago

Sir , wouldn't it be better to say this way that since ⌊ 2 x ⌋ ≥ 2 ⌊ x ⌋ \lfloor{2x}\rfloor \geq 2\lfloor{x}\rfloor

⇒ 1 ⌊ x ⌋ + 1 ⌊ 2 x ⌋ ≤ 1 ⌊ x ⌋ + 1 2 ⌊ x ⌋ \Rightarrow \dfrac1{\lfloor{x}\rfloor} + \dfrac1{\lfloor{2x}\rfloor} \leq \dfrac1{\lfloor{x}\rfloor} + \dfrac1{2\lfloor{x}\rfloor} than to say that 'for integer x x ....'

and then proceed similarly??

Ankit Kumar Jain - 4 years, 2 months ago

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