2016 AMC 10B Problem 25

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}

Only the fractional part of

x

matters

Count distinct breakpoints

\frac{i}{k}

for

2\le k\le10

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

\frac{i}{k}

, where

2\le k\le10

and

1\le i\lt k

. The number of distinct such fractions in

(0,1)

is

\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,

f

assumes

31+1=32

distinct values.

Thus, the correct answer is A.

Problem 25 in other years