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# Practice Paper 3 Question 15

Find all values of $$x\geq0$$ in terms of $$k$$ that satisfy $$\lfloor kx \rfloor = (k+1)\lfloor x \rfloor,$$ where $$k>0$$ is an integer and $$\lfloor r \rfloor$$ is the greatest integer less than or equal to $$r.$$ You may wish to consider $$k=2$$ first.

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## Warm-up Questions

1. Let $$f(x)=x^3 - 3x^2 - 3x + 1$$ and $$g(x) = x+1.$$ For which values of $$x$$ is $$g(x) > f(x)?$$
2. Find the set of solutions for $$x$$ that satisfies $$\lfloor \frac{x}{2} \rfloor = 4.$$

## Hints

• Hint 1
By rewriting $$x,$$ how can one simplify the equation?
• Hint 2
... in particular, let $$x = n+f,$$ where $$n$$ is the integer part and $$f$$ is the fractional part $$(0\le f<1).$$
• Hint 3
Consider the relation between $$\lfloor n+f \rfloor$$ and $$n.$$
• Hint 4
After simplifying, try considering the different possible cases for $$n.$$
• Hint 5
... and the corresponding values of $$f?$$

## Solution

In this solution we use the notation $$x\in[a,b)$$ to mean $$a \leq x < b.$$

For the general case, let $$x=n+f$$ where $$n$$ is the integer part and $$f\in[0,1)$$ is the fractional part. We have $$\lfloor kn+kf \rfloor=(k+1)\lfloor n+f \rfloor$$ which gives $$kn+\lfloor kf \rfloor=(k+1)n$$ and finally $$\lfloor kf \rfloor=n.$$ This means that $$n$$ can only take the values $$n=0,1,\ldots,k-1.$$ Taking these in turn we get:

• For $$n=0,$$ $$f\in[0,\frac{1}{k})$$ and so $$x\in[0,\frac{1}{k}).$$
• For $$n=1,$$ $$f\in[\frac{1}{k},\frac{2}{k})$$ and so $$x\in[1+\frac{1}{k},1+\frac{2}{k}).$$
• $$\cdots$$
• For $$n=r,$$ $$f\in[\frac{r}{k},\frac{r+1}{k})$$ and so $$x\in[r+\frac{r}{k},r+\frac{r+1}{k}).$$
• $$\cdots$$

In the end, we take the union of all these sets up to $$n=k-1,$$ which is formally written as $$x\in\bigcup\limits_{n=0}^{k-1} \big[n+\frac{n}{k}, n+\frac{n+1}{k}\big).$$

Note: this notation is not required for full credit.

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