MathDB
f(x)=x+1+[x]

Source: Turkish NMO 1st Round - 2013 - Problem 35

April 19, 2013
floor functionalgebra proposedalgebra

Problem Statement

What is the least positive integer nn such that f(f(…f⏞21 times(n)))=2013\overbrace{f(f(\dots f}^{21 \text{ times}}(n)))=2013 where f(x)=x+1+⌊x⌋f(x)=x+1+\lfloor \sqrt x \rfloor? (⌊a⌋\lfloor a \rfloor denotes the greatest integer not exceeding the real number aa.)
<spanclass=′latex−bold′>(A)</span> 1214<spanclass=′latex−bold′>(B)</span> 1202<spanclass=′latex−bold′>(C)</span> 1186<spanclass=′latex−bold′>(D)</span> 1178<spanclass=′latex−bold′>(E)</span> None of above <span class='latex-bold'>(A)</span>\ 1214 \qquad<span class='latex-bold'>(B)</span>\ 1202 \qquad<span class='latex-bold'>(C)</span>\ 1186 \qquad<span class='latex-bold'>(D)</span>\ 1178 \qquad<span class='latex-bold'>(E)</span>\ \text{None of above}
Back to ProblemsView on AoPS