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2016 AMC 10B Problem 25

Problem 25 of 25HarderAlgebraNumber Theory

Let f(x)=k=210(kxkx),f(x)=\sum_{k=2}^{10}(\lfloor kx \rfloor -k \lfloor x \rfloor), where r\lfloor r \rfloor denotes the greatest integer less than or equal to r.r. How many distinct values does f(x)f(x) assume for x0?x \ge 0?

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Solution

Write x=x+tx=\lfloor x\rfloor+t, where 0t<10\le t\lt1. Then kxkx=kt,\lfloor kx\rfloor-k\lfloor x\rfloor=\lfloor kt\rfloor, so f(x)f(x) depends only on the fractional part tt. The value of ff changes only when tt crosses a fraction ik\frac{i}{k}, where 2k102\le k\le10 and 1i<k1\le i\lt k. The number of distinct such fractions in (0,1)(0,1) is φ(2)+φ(3)++φ(10)=1+2+2+4+2+6+4+6+4=31. \begin{aligned} &\varphi(2)+\varphi(3) \\ &\quad {}+\cdots+\varphi(10) \\ &=1+2+2+4+2 \\ &\quad {}+6+4+6+4 \\ &=31. \end{aligned} Including the initial value before the first breakpoint, ff assumes 31+1=3231+1=32 distinct values. Thus, the correct answer is A.

More practice

Concepts: floor and ceiling functions · Euler’s Totient Function

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.