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How can I calculate the number of perfect cubes among the first $4000$ positive integers?

Is there any trick to solving such questions?

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  • $\begingroup$ ^I did, thanks for reminding me. $\endgroup$
    – Iti Shree
    Commented Apr 29, 2017 at 16:59

4 Answers 4

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If you live in the world of base-2 geekdom, simply note that $2^{12}$ a.k.a. $16^3$ is 4K, or 4096. This is obviously too large. The barest mental math estimation will verify that $15^3 < 4000$. Done.

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You could find the largest cube $\le 4000$.

$\sqrt[3]{4000} \approx 15.9$, so there are $15$ perfect cubes among the first $4000$ positive integers.

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    $\begingroup$ I think it's better to evaluate $15^3$ and $16^3$ and thus conclude there are $15$ perfect cubes among the first $4000$ - assuming no access to calculator. $\endgroup$
    – mrnovice
    Commented Apr 25, 2017 at 14:40
  • $\begingroup$ @ItiShree You're welcome! $\endgroup$
    – Ovi
    Commented Apr 25, 2017 at 14:44
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If you didn't have a calculator and needed to work out what the largest integer $x$ was such that $x^3\leq4000$ without just computing $\sqrt[3]{4000}$, then you could estimate it - $10^3=1000$, $20^3=8000$ so $10<x<20$. Then keep narrowing it down, e.g. go halfway $15^3=225\cdot15=2250+1125=3375$, and $16^3=256\cdot16=2560+1536>4000$.

So there are $15$.

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  • $\begingroup$ Yeah I can't use calculator, thanks. $\endgroup$
    – Iti Shree
    Commented Apr 25, 2017 at 14:43
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    $\begingroup$ @ItiShree If you have any such restrictions (some method or tool is forbidden), please say so in the question. That helps others answer the question you actually have. Not all of the answerers guessed your intentions correctly. $\endgroup$ Commented Apr 25, 2017 at 21:41
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Is there any trick to find such questions?

Google, I would suppose. I think you meant to say "to find the answers to such questions." And yes, there is, though to me it seems too elementary to call it a "trick." To the find how many perfect $k$-th powers there are among the first $n$ positive integers, you just have to compute $\lfloor \root k \of n \rfloor$. (Note that if $k$ is even, you don't strictly need to specify "positive," but your answer might be off by $1$ from the answer key if it's not clear whether $0^k$ is meant to be included.)

How to calculate number of perfect cubes among the among the first $4000$ positive integers?

We see that $\root 3 \of {4000} \approx 15.874$, and then we verify that $15^3 = 3375 < 4000$ and $16^3 = 4096 > 4000$. Therefore the cubes you're looking for are $$1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375$$ (the first sixteen listed in Sloane's A000578, but with $0$ omitted).

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