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2016 AMC 10B Problem 25

Problem 25 of 25HarderAlgebraNumber Theory

Let f(x)=∑k=210(⌊kx⌋−k⌊x⌋),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 x≥0?x \ge 0?

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Solution

Write x=⌊x⌋+tx=\lfloor x\rfloor+t, where 0≤t<10\le t\lt1. Then ⌊kx⌋−k⌊x⌋=⌊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 2≤k≤102\le k\le10 and 1≤i<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.
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Tagged: floor and ceiling functions · Euler’s Totient Function

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