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Olympiad workbook
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Olympiad workbook
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IOQM
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Functions

Olympiad workbook · IOQM · Functions

For any real number x , let [x] denotes the integer part of x ;\ x\...

For any real number xx, let [x][x] denotes the integer part of x;{x}x ;\{x\} be the fractional part of x({x}=x−∣x∣)x (\{x\}=x-|x|). Let AA denote the set of all real numbers xx satisfying {x}=x+[x]+[x+(1/2)]20\{x\}=\frac{x+[x]+[x+(1 / 2)]}{20} If SS is the sum of all numbers in AA, find [S][S]

Correct answer

21

Step-by-step solution

Let, x=l+fx=l+f ∴\therefore given equation reduces to, f=\frac{I+f+I+I+\left[f+\frac{1}{2}\right]}{20}  (l) f∈[0,12)[f+12]=0∴19f=3l⇒f=3l190,<3l19<12⇒l∈[0,196)⇒l=0,1,2,3∴f=0,319,619,919∴x=0,2219,4419,6619\begin{array}{ll} \text { (l) } & f \in\left[0, \frac{1}{2}\right) \\ & {\left[f+\frac{1}{2}\right]=0} \\ \therefore & 19 f=3 l \Rightarrow f=\frac{3 l}{19} \\ & 0, < \frac{3 l}{19} < \frac{1}{2} \\ \Rightarrow & l \in\left[0, \frac{19}{6}\right) \Rightarrow l=0,1,2,3 \\ \therefore & f=0, \frac{3}{19}, \frac{6}{19}, \frac{9}{19} \\ \therefore & x=0, \frac{22}{19}, \frac{44}{19}, \frac{66}{19} \end{array}  (II) f∈[12,1][f+12]=120f=3l+1+ff=3l+11912≤3l+119<1⇒176≤l<6⇒l=3,4,5∴f=1019,1319,1619∴x=6719,8919,11119∴S=39919[S]=21\begin{array}{ll}\text { (II) } & f \in\left[\frac{1}{2}, 1\right] \\ & {\left[f+\frac{1}{2}\right]=1} \\ & 20 f=3 l+1+f \\ & f=\frac{3 l+1}{19} \\ & \frac{1}{2} \leq \frac{3 l+1}{19} < 1 \Rightarrow \frac{17}{6} \leq l < 6 \Rightarrow \quad l=3,4,5 \\ \therefore & f=\frac{10}{19}, \frac{13}{19}, \frac{16}{19} \\ \therefore & x=\frac{67}{19}, \frac{89}{19}, \frac{111}{19} \\ \therefore & S=\frac{399}{19} \\ & {[S]=21}\end{array}
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