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  1. Smallest n with many solutions ●●●●○ 100 ptsLet f(x)=∣2{x}−1∣f(x)=|2\{x\}-1|, where {x}\{x\} denotes the fractional part of xx. Find the smallest positive integer nn such that the equation nf(xf(x))=xn f(x f(x))=x has at least 20122012 real solutions.