Find all the positive perfect cubes that are not divisible by $10$ such that the number obtained by erasing the last three digits is also a perfect cube.
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Tags: geometry, number theory proposed, number theory
pco
31.10.2010 11:13
WakeUp wrote: Find all the positive perfect cubes that are not divisible by $10$ such that the number obtained by erasing the last three digits is also a perfect cube. So $(10x)^3+a=y^3$ with $a\in\{1,2,...,999\}$ So $1000>a=y^3-(10x)^3\ge (10x+1)^3-(10x)^3$ So $300x^2+30x-999<0$ So $x<1.7755...$ So : either $x=0$ and $a=y^3$ and the solutions $1,8,27,64,125,216,343,512,729$ either $x=1$ and $a=y^3-1000$ and the solutions $1331,1728$ Hence the answer : $\boxed{1,8,27,64,125,216,343,512,729,1331,1728}$
Binomial-theorem
28.04.2012 23:37
Let $N$ be the number obtained by erasing the last $3$ digits of the number. $N$ is a perfect cube, and we also desire $1000N+100a+10b+c$ to be a perfect cube. Therefore, let $N=n^3$ to give us: $1000n^3+100a+10b+c$ needs to be a perfect cube. Let it be $m^3$. Note that we need $(10n)^3+100a+10b+c=m^3$ giving us $100a+10b+c=m^3-(10n)^3=(m-10n)(m^2+10mn+100n^2)$. Note that $m\neq 10n$ since that would imply $c=0$ which isn't allowed. Hence $m-10n\ge 1$ and therefore since $a\le 9$ (from it being a base $10$ digit), we have $n=1, 2, 3$.
$n=1$ gives $100a+10b+c=(m-10)(m^2+10m+100)$. $m=11$ gives $121+110+100=331$ therefore $1331$ works. $m=12$ gives $2(144+120+100)=2(364)=728$. Therefore $1728$ works also. $m=13$ gives $3(169+130+100)$ and since $169+130+100>334$, we are not going to have this being able to work.
$n=2$ gives us $100a+10b+c=(m-20)(m^2+20m+400)$. $m=21$ gives us $1(21^2+20*21+400)$ which is greater than $1000$ therefore this can't work also.
$ n=3$ gives us $100a+10b+c=(m-30)(m^2+30m+900)$. $m=31$ already gives us $100a+10b+c>1000$ which we can't have again.
Also, all one to three digit perfect cubes work, giving $n=\boxed{1331, 1728}$.
Edit: Good point @below post. Fixed
mavropnevma
28.04.2012 23:58
Erasing the last three digits of $\boxed{1,8,27,64,125,216,343,512,729}$ leaves nothing; we do not write, for example, $27 = 0027$, so that by erasing its last three digits we are left with the perfect cube $0$. So I think the problem was meant to have as answer $\boxed{1331,1728}$.
aar010123
14.11.2019 23:13
mavropnevma wrote: Erasing the last three digits of $\boxed{1,8,27,64,125,216,343,512,729}$ leaves nothing; we do not write, for example, $27 = 0027$, so that by erasing its last three digits we are left with the perfect cube $0$. So I think the problem was meant to have as answer $\boxed{1331,1728}$. 1000 too
franzliszt
15.11.2019 00:25
WakeUp wrote: Find all the positive perfect cubes that are not divisible by $10$ such that the number obtained by erasing the last three digits is also a perfect cube.