Find the maximum value of the quotient

Given a real number $x,$ let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x.$ For a certain integer $k,$ there are exactly $70$ positive integers $n_{1}, n_{2}, \ldots, n_{70}$ such that $k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{1}}\rfloor = \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor$ and $k$ divides $n_{i}$ for all $i$ such that $1 \leq i \leq 70.$ Find the maximum value of $\frac{n_{i}}{k}$ for $1\leq i \leq 70.$

It is an interesting problem. I would appreciate if I could have HINTS ONLY.

So: $n_i \equiv 0 \pmod{k}$, and so: $n_i = r_i(k)$ for some $r_i$.

Also it is seen that:

$t - 1 < k \le k$ using the floor function definition. (definition for $t$) is below:

Let $\lfloor n_i \rfloor = t$, I need to find an interval such that: $f(x) = \lfloor \sqrt[3]{x} \rfloor$ is constant for seventy integer $x$ values.

I had several attempts , but I cant do it.

I tried to set:

$(x +1)^3 - x^3 = 70$, but that doesnt give an integer $x$.

You have to look for a number $k$ such that between $k^3$ and $(k+1)^3$ there are exactly $70$ multiples of $k$.
Since $k^3$ and $(k+1)^3-1$ are multiples of $k$ this gives $$[(k+1)^3-1]-k^3=69k$$
• why doesnt: $(k + 1)^3 - k^3 = 70k - 1$ work – Amad27 Aug 24 '15 at 15:44
• Then you'll have $71$ numbers. Note $k^2.k, (k^2+1)k,...(k^2+69)k$ are the $70$ multiples and the last one should be $=(k+1)^3-1$. – Macavity Aug 24 '15 at 16:01
• @Macavity, I dont understand the equation. We look for a number $k$ such that in between there are only $70$ $k$ multiplies. The first is: $n_0 = k^3$ and the last is obviously, $n_l = (k + 1)^3 - 1$. Why is it: $(k +1^3 - 1 - k^3 = 69k$ and not: $(k +1^3 - 1) - k^3 = 70k$ – Amad27 Aug 24 '15 at 16:20
• @Amad27 Here is another way to think - the multiples form an A.P. with common difference $k$. The first term is $k^3$ and so the $70$th term is $k^3+(70-1)k$. This must be equal to $(k+1)^3-1$. – Macavity Aug 24 '15 at 16:25