Inspired from Mistakes Give Rise To Problems...

Algebra Level 5

In maths , there exists the following property

⌊ ⌊ x 2 ⌋ + ⌊ x 2 ⌋ ⌋ = ⌊ x 2 ⌋ + ⌊ x 2 ⌋ \left\lfloor \left\lfloor \frac { x }{ 2 } \right\rfloor +\left\lfloor { x }^{ 2 } \right\rfloor \right\rfloor =\left\lfloor \frac { x }{ 2 } \right\rfloor +\left\lfloor { x }^{ 2 } \right\rfloor

but if you do

{ { x 2 } + { x 2 } } = { x 2 } + { x 2 } \left\{ \left\{ \frac { x }{ 2 } \right\} +\left\{ { x }^{ 2 } \right\} \right\} =\left\{ \frac { x }{ 2 } \right\} +\left\{ { x }^{ 2 } \right\}

then this is a big Mistake!!!

but the above expression is true for some set k where k contains all possible values of x from -2 <x<2 and sum of end points of all intervals of k can be expressed as s

Find ⌊ 100 { s } ⌋ \left\lfloor 100\left\{ s \right\} \right\rfloor

Details and assumptions

1.{x} is fractional part of x, which we will define as x − ⌊ x ⌋ x - \lfloor x \rfloor , even for negative numbers.

2. ⌊ ⌋ \left\lfloor \quad \right\rfloor is greatest integer less than or equal to x.

3.For example if your answer is

k = [ − 3 2 , − 1 ) ∪ ( 2 , 2 ) = \left[ -\frac { 3 }{ 2 } ,-1 \right) \cup \left( \sqrt { 2 } ,2 \right)

then s = − 3 2 − 1 + 2 + 2 -\frac { 3 }{ 2 } -1+\sqrt { 2 } +2

4.Consider both cases whethere end points are open ( ) or closed [ ]

5.This is my second problem ever

6.This problem is inspired from the set Mistakes Give Rise to Problems


The answer is 76.

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Ayush Garg by Ayush Garg , 24, India

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1 solution

Ayush Garg
Aug 2, 2014

Aditya Raut I`ve not added the problem to your set.If you like the problem, then feel free to add it to your set.

I`m just posting a brief solution (not explaining anything in detail)

Note that above equality will hold if RHS <1

So, we equate it to 1 to get

x = 17 − 1 4 , 33 − 1 4 , 3 2 , 65 − 1 4 , − 1 2 , − 17 − 1 4 , − 33 − 1 4 x\quad =\quad \frac { \sqrt { 17 } -1 }{ 4 } ,\frac { \sqrt { 33 } -1 }{ 4 } ,\frac { 3 }{ 2 } ,\frac { \sqrt { 65 } -1 }{ 4 } ,-\frac { 1 }{ 2 } ,\frac { -\sqrt { 17 } -1 }{ 4 } ,\frac { -\sqrt { 33 } -1 }{ 4 }

Note that at x = 17 − 1 4 x\quad =\quad \frac { \sqrt { 17 } -1 }{ 4 } , RHS=1 and at x=1,RHS = 1 2 < 1 \frac { 1 }{ 2 } <1 ,

Therefore, when x 2 { x }^{ 2 } =0, an inteval will either start or end (except for x=0)

so we get x = 1 , 2 , 3 , − 2 , − 1 , − 2 , − 3 , x\quad =\quad 1,\sqrt { 2 } ,\sqrt { 3 } ,-2,-1,-\sqrt { 2 } ,-\sqrt { 3 } ,

Adding them we get

s = 17 − 1 4 + 33 − 1 4 + 3 2 + 65 − 1 4 − 1 2 + − 17 − 1 4 + − 33 − 1 4 + 1 + 2 + 3 − 2 − 1 − 2 − 3 s\quad =\quad \frac { \sqrt { 17 } -1 }{ 4 } +\frac { \sqrt { 33 } -1 }{ 4 } +\frac { 3 }{ 2 } +\frac { \sqrt { 65 } -1 }{ 4 } -\frac { 1 }{ 2 } +\frac { -\sqrt { 17 } -1 }{ 4 } +\frac { -\sqrt { 33 } -1 }{ 4 } +1+\sqrt { 2 } +\sqrt { 3 } -2\quad -1\quad -\sqrt { 2 } \quad -\sqrt { 3 }

s= 65 − 9 4 \frac { \sqrt { 65 } -9 }{ 4 } = -0.2344....

{s} = 0.765......

100{s} = 76.5.......

⌊ 100 { s } ⌋ \left\lfloor 100\left\{ s \right\} \right\rfloor = 76

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I`m just posting a brief solution (not explaining anything in detail)

Note that above equality will hold if RHS <1

So, we equate it to 1 to get

x = 17 − 1 4 , 33 − 1 4 , 3 2 , 65 − 1 4 , − 1 2 , − 17 − 1 4 , − 33 − 1 4 x\quad =\quad \frac { \sqrt { 17 } -1 }{ 4 } ,\frac { \sqrt { 33 } -1 }{ 4 } ,\frac { 3 }{ 2 } ,\frac { \sqrt { 65 } -1 }{ 4 } ,-\frac { 1 }{ 2 } ,\frac { -\sqrt { 17 } -1 }{ 4 } ,\frac { -\sqrt { 33 } -1 }{ 4 }

Note that at x = 17 − 1 4 x\quad =\quad \frac { \sqrt { 17 } -1 }{ 4 } , RHS=1 and at x=1,RHS = 1 2 < 1 \frac { 1 }{ 2 } <1 ,

Therefore, when x 2 { x }^{ 2 } =0, an inteval will either start or end (except for x=0)

so we get x = 1 , 2 , 3 , − 2 , − 1 , − 2 , − 3 , x\quad =\quad 1,\sqrt { 2 } ,\sqrt { 3 } ,-2,-1,-\sqrt { 2 } ,-\sqrt { 3 } ,

Adding them we get

s = 17 − 1 4 + 33 − 1 4 + 3 2 + 65 − 1 4 − 1 2 + − 17 − 1 4 + − 33 − 1 4 + 1 + 2 + 3 − 2 − 1 − 2 − 3 s\quad =\quad \frac { \sqrt { 17 } -1 }{ 4 } +\frac { \sqrt { 33 } -1 }{ 4 } +\frac { 3 }{ 2 } +\frac { \sqrt { 65 } -1 }{ 4 } -\frac { 1 }{ 2 } +\frac { -\sqrt { 17 } -1 }{ 4 } +\frac { -\sqrt { 33 } -1 }{ 4 } +1+\sqrt { 2 } +\sqrt { 3 } -2\quad -1\quad -\sqrt { 2 } \quad -\sqrt { 3 }

s= 65 − 9 4 \frac { \sqrt { 65 } -9 }{ 4 } = -0.2344....

{s} = 0.765......

100{s} = 76.5.......

⌊ 100 { s } ⌋ \left\lfloor 100\left\{ s \right\} \right\rfloor = 76

Ayush Garg - 6 years, 10 months ago

the fractional part of a negative number has two possible values , that is, no unique universal definition, according to wikipedia. there is no clue in the question that tell us what definition we should use ...

Hasan Kassim - 6 years, 10 months ago

For -tive x, it is always true. When the expression on the right is less than 1, the equation is true. So the limit when it is just false is x^2+x/2 - 1 = 0. this gives x=.78.-->{s}=.78.

Niranjan Khanderia - 6 years, 9 months ago

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