# 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.

## Related topics

## Warm-up Questions

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

## Hints

- Hint 1By 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 3Consider the relation between \(\lfloor n+f \rfloor\) and \(n.\)
- Hint 4After 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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