MathDB

Problem 11

Part of 2019 LIMIT Category B

Problems(2)

counting with triangle inequality

Source: LIMIT 2019 CBS1 P11

4/28/2021
Let S={1,2,…,10}S=\{1,2,\ldots,10\}. Three numbers are chosen with replacement from SS. If the chosen numbers denote the lengths of sides of a triangle, then the probability that they will form a triangle is: <spanclass=′latex−bold′>(A)</span> 101200<span class='latex-bold'>(A)</span>~\frac{101}{200} <spanclass=′latex−bold′>(B)</span> 99200<span class='latex-bold'>(B)</span>~\frac{99}{200} <spanclass=′latex−bold′>(C)</span> 12<span class='latex-bold'>(C)</span>~\frac12 <spanclass=′latex−bold′>(D)</span> 110200<span class='latex-bold'>(D)</span>~\frac{110}{200}
combinatoricsgeometrytriangle inequalityinequalities
floor function, size comparison

Source: LIMIT 2019 CBS2 P11

4/28/2021
⌊(1⋅2+2⋅22+…+100⋅2100)⋅9−901⌋=?\left\lfloor\left(1\cdot2+2\cdot2^2+\ldots+100\cdot2^{100}\right)\cdot9^{-901}\right\rfloor=?
algebrafloor functionfunctionBounding