Counting Conundrum

Relevant wiki: Trigonometric Functions Problem Solving

Let $f(x)=|x \sin(\pi x)|$ . Note that counting the number of $x$ such that $f(x)$ is an integer is the same as the original problem, since

$x \sin(\pi x) \in \mathbb{Z} \rightarrow |x \sin(\pi x)| \in \mathbb{N} \subset \mathbb{Z}$

Also, note the following properties of $f(x)$ :

• $g(x)=x$ is tangent to $f(x)$ if and only if $x=n+\frac{1}{2}$ for some non-negative integer $n \quad (1)$

• $f(x)\leq g(x)$ for all $x \geq 0 \quad (2)$

• $f(n)=0$ for all integers $n \quad (3)$

Putting it all together, for $n \leq x \leq n+1$ , we have that $f(x)$ begins at $0$ (3), goes up to some maximum $n<M<n+1$ (1,2) and then returns back to $f(x)=0$ when $x=n+1$ (3)

From here, we know that if $n$ is a non-negative integer, for each interval $n < x < n+1$ we get that $f(x)$ crosses $2n$ integers, and that $f(n)$ is also an integer.

Hence we get the $2018$ integers from $0$ to $2017$ inclusive, as well as

$\sum_{n=0}^{2016} 2n = 2016(2017) = 4066272$

The total is $4066272+2018=\boxed{4068290}$ .

Note: for completeness, we should show that $f(x)$ between $n<x<n+1$ is strictly increasing up until a maximum, and then strictly decreasing until it once again reaches $0$ . I can't think of an easy way to show this without using calculus, however.

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Bonus:

We get $n+1$ from when $x$ is an integer, and the rest come from the sum in our solution, which evaluates to $n(n-1)$

$n+1+(n)(n-1)=\boxed{n^{2}+1}$

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A graph to better understand solution:

Blue curve: $f(x)=x$

Red curve: $f(x)=\left| x \sin (\pi x) \right|$

Relevant wiki: Trigonometric Functions Problem Solving

For $n > 0$ consider the function $f(x) = |x\sin(\pi x)|$ on the interval $\langle n-1,n]$ .

On the first half of the interval, the function increases monotonously from $f(n-1) = 0$ to $f(n-\tfrac12) = n-\tfrac12$ . Thus $f\langle n-1,n-\tfrac12] = \langle 0, n-\tfrac12]$ On the second half of the interval, the function decreases monotonously from $f(n-\tfrac12) = n-\tfrac12$ to $f(n) = 0$ . Thus $f\langle n-\tfrac12,n] = [0, n-\tfrac12\rangle$ The first image contains $n-1$ integers; the second image, $n$ integers. Since the function is strictly monotonous on both parts, it is injective on each half interval, so that:

The interval $\langle n-1,n]$ contains $2n - 1$ values for $x$ so that $f(x)$ is an integer.

Removing the absolute value in the definition of $f(x)$ does not change this conclusion for obvious reasons.

Thus on the interval $\langle 0, N]$ we obtain $\sum_{n=1}^N 2n-1 = 1 + 3 + 5 + \dots = N^2$ Finally, the problem asks for the number of integers in the image of the closed interval, $[0, N]$ . We must therefore add one since $f(0) =0$ . The answer is $\boxed{N^2 + 1} = 2017^2 + 1 = \boxed{4\,068\,290}.$

Relevant wiki: Trigonometric Functions Problem Solving

First notice that $x\sin(\pi x)$ is zero whenever $sin(\pi x)$ is zero. This occurs when x is an integer so 2018 times from 0 to 2017 inclusive.

Then notice that $x\sin(\pi x)$ is equal to x whenever $sin(\pi x)$ is equal to 1 or -1. This occurs between each zero so the graph of $f(x)=x\sin(\pi x)$ must have two places with a height of -1 somewhere between x=1 and x=2. Then four integer intersections somewhere between x=2 and x=3. Then six integer intersections somewhere between x=3 and x=4. This pattern continues. To count the non-zero intersections we just need the sum: 2+4+6+8+...+4032. So, 2018+ $\sum_{n=1}^{2016}2n$ =4068290