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Can we find the number of perfect cubes in a range.

For eg if L = 7 and R = 220 then count will be 5 because 8, 27, 64, 125, 216 are the perfect squares in

the range L to R. Can we make a formula for this?

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Yes, we can find a formula.

I am assuming $L\leq R$, and we include $L$ and $R$ themselves in the range. Take the cube roots of $L$ and $R$. Round $\sqrt[3]L$ up and $\sqrt[3]R$ down. Take their difference, and add $1$, and you're done. In symbols: $$ \left\lfloor \sqrt[3]R\right\rfloor - \left\lceil\sqrt[3]L\right\rceil + 1 $$ (Changing whether $L$ and $R$ themselves count, should they happen to be cubes, will change whether we round up or down, and exactly which number we add at the end. For instance, if $R$ counts, but $L$ does not, then we round them both down, and do not add anything at the end.)

To see how this formula works, let's take your example numbers. Then $\sqrt[3]L \approx 1.9$. Rounding this up gives the lowest number whose cube is above $L$, namely $2$ (as $2^3$ is large enough, but $1^3$ isn't).

We also have $\sqrt[3]{R}\approx 6.04$. Rounding this down gives the largest number whose cube is below $R$, namely $6$ (as $6^3$ is small enough, but $7^3$ is too large).

So all the integers from $2$ to $6$ have cubes in our range, and no other integers do. And from $2$ to $6$ there are $$ 6-2 + 1 = 5 $$ integers.

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$$ \lfloor \sqrt[3]{R} \rfloor - \lceil \sqrt[3]{L} \rceil + 1 = \lfloor \sqrt[3]{220} \rfloor - \lceil \sqrt[3]{7} \rceil + 1 =6-2+1=5 $$

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