I'm Floored!

Algebra Level 4

$\large \dfrac{\lfloor x \rfloor}{x} = \dfrac {9}{10}$

Find the largest real $x$ such that the above equation is satisfied. If your answer is of the form of $\dfrac ab$ , where $a$ and $b$ are coprime positive integers, find $a+b$ .

Try more here

The answer is 89.

2 solutions

Neelesh Vij
Jan 29, 2016

We can write the equation as $10\times \lfloor x \rfloor = 9x$

$\rightarrow10\times \lfloor x \rfloor = 9( \lfloor x \rfloor + [x])$ ; where $[x]$ is fractional part of $x$

$\rightarrow \lfloor x \rfloor = 9[x]$

Now we can see that LHS is an integer so RHS should be an integer

$\therefore [x] = \frac19 , \frac29 , \frac 39, ..... \frac89$

For largest root we take $[x] = \frac 89$

So we get $x = \lfloor x \rfloor + [x] = 8 + \frac89 = \frac{80}{9}$

So our answer is $80 + 9 = \boxed{89}$

nice problem, just one remark that the greatest integer less than or equal to $x$ is written as $\lfloor x \rfloor$

Abdeslem Smahi - 5 years, 4 months ago

Nice problem with a nice solution. Thanks!

Sandeep Bhardwaj - 5 years, 3 months ago

Exactly Same Way.

Kushagra Sahni - 5 years, 3 months ago

Ajinkya Shivashankar
Jan 29, 2016

x-1<[x]<=x

[x]=9x/10

solving

[x]<9

[x]max=8

xmax=8*10/9=80/9

I'm Floored!

The answer is 89.

2 solutions

0 pending reports