Question

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

$\ 32$

$\ 36$

$\ 45$

$\ 46$

$\ \text{infinitely many}$

Solution:

Write $x=\lfloor x\rfloor+t$ , where $0\le t\lt1$ . Then $\lfloor kx\rfloor-k\lfloor x\rfloor=\lfloor kt\rfloor,$ so $f(x)$ depends only on the fractional part $t$ .

The value of $f$ changes only when $t$ crosses a fraction $i/k$ , where $2\le k\le10$ and $1\le i\lt k$ . The number of distinct such fractions in $(0,1)$ is $\varphi(2)+\varphi(3)+\cdots+\varphi(10)=1+2+2+4+2+6+4+6+4=31.$

Including the initial value before the first breakpoint, $f$ assumes $31+1=32$ distinct values.

Thus, the correct answer is A.

2016 AMC 10B Problem 25

Solution:

Problem 25 in Other Years