Dividing Hypercubes into $n$ smaller Hypercubes












1












$begingroup$


Name a positive integer $n$ nice if a square can be divided into $n$ smaller squares. The smaller squares do not need to be of the same size. Since you can always divide a square into $4$ smaller squares it immediately follows, that if $n$ is nice $n+3$ has to be aswell. Since $6, 7$ and $8$ are nice all natural numbers greater than $8$ have to be nice.



This got me thinking about the same problem in higher Dimensions.
Let $n_d$ be nice if it divides a Hypercube in $d$ Dimensions into $n_d$ smaller Hypercubes.



Does for all Dimensions $d$ exist a $n_d$ such that all numbers greater than $n_d$ are nice? Is there a simple way to determine wether a number is nice in $d$ Dimensions or not?






combinatorics geometry logic proof-writing modular-arithmetic







share|cite|improve this question









$endgroup$












  • $begingroup$
    By the same argument you get that in dimension $d$, if $n$ is nice also $n+2^d-1$ is nice. Also, the only number smaller than $2^d$ that is nice is $1$. But this doesn't help a lot, I guess.
    $endgroup$
    – Servaes
    Dec 13 '18 at 19:33










  • $begingroup$
    How do you divide a square in $6$ squares?
    $endgroup$
    – ajotatxe
    Dec 13 '18 at 19:36






  • 2




    $begingroup$
    @ajotatxe: Divide a $3 times 3$ square into one $2 times 2$ and five $1 times 1$.
    $endgroup$
    – Robert Israel
    Dec 13 '18 at 19:40










  • $begingroup$
    $8$ is a bit harder: see here
    $endgroup$
    – Robert Israel
    Dec 13 '18 at 20:05






  • 1




    $begingroup$
    @RobertIsrael Wouldn't a $4times4$ divided into one $3times3$ and seven $1times1$ squares also be a dissection into $8$ squares?
    $endgroup$
    – David K
    Dec 13 '18 at 20:17


















1












$begingroup$


Name a positive integer $n$ nice if a square can be divided into $n$ smaller squares. The smaller squares do not need to be of the same size. Since you can always divide a square into $4$ smaller squares it immediately follows, that if $n$ is nice $n+3$ has to be aswell. Since $6, 7$ and $8$ are nice all natural numbers greater than $8$ have to be nice.



This got me thinking about the same problem in higher Dimensions.
Let $n_d$ be nice if it divides a Hypercube in $d$ Dimensions into $n_d$ smaller Hypercubes.



Does for all Dimensions $d$ exist a $n_d$ such that all numbers greater than $n_d$ are nice? Is there a simple way to determine wether a number is nice in $d$ Dimensions or not?






combinatorics geometry logic proof-writing modular-arithmetic







share|cite|improve this question









$endgroup$












  • $begingroup$
    By the same argument you get that in dimension $d$, if $n$ is nice also $n+2^d-1$ is nice. Also, the only number smaller than $2^d$ that is nice is $1$. But this doesn't help a lot, I guess.
    $endgroup$
    – Servaes
    Dec 13 '18 at 19:33










  • $begingroup$
    How do you divide a square in $6$ squares?
    $endgroup$
    – ajotatxe
    Dec 13 '18 at 19:36






  • 2




    $begingroup$
    @ajotatxe: Divide a $3 times 3$ square into one $2 times 2$ and five $1 times 1$.
    $endgroup$
    – Robert Israel
    Dec 13 '18 at 19:40










  • $begingroup$
    $8$ is a bit harder: see here
    $endgroup$
    – Robert Israel
    Dec 13 '18 at 20:05






  • 1




    $begingroup$
    @RobertIsrael Wouldn't a $4times4$ divided into one $3times3$ and seven $1times1$ squares also be a dissection into $8$ squares?
    $endgroup$
    – David K
    Dec 13 '18 at 20:17
















1












1








1





$begingroup$


Name a positive integer $n$ nice if a square can be divided into $n$ smaller squares. The smaller squares do not need to be of the same size. Since you can always divide a square into $4$ smaller squares it immediately follows, that if $n$ is nice $n+3$ has to be aswell. Since $6, 7$ and $8$ are nice all natural numbers greater than $8$ have to be nice.



This got me thinking about the same problem in higher Dimensions.
Let $n_d$ be nice if it divides a Hypercube in $d$ Dimensions into $n_d$ smaller Hypercubes.



Does for all Dimensions $d$ exist a $n_d$ such that all numbers greater than $n_d$ are nice? Is there a simple way to determine wether a number is nice in $d$ Dimensions or not?






combinatorics geometry logic proof-writing modular-arithmetic







share|cite|improve this question









$endgroup$




Name a positive integer $n$ nice if a square can be divided into $n$ smaller squares. The smaller squares do not need to be of the same size. Since you can always divide a square into $4$ smaller squares it immediately follows, that if $n$ is nice $n+3$ has to be aswell. Since $6, 7$ and $8$ are nice all natural numbers greater than $8$ have to be nice.



This got me thinking about the same problem in higher Dimensions.
Let $n_d$ be nice if it divides a Hypercube in $d$ Dimensions into $n_d$ smaller Hypercubes.



Does for all Dimensions $d$ exist a $n_d$ such that all numbers greater than $n_d$ are nice? Is there a simple way to determine wether a number is nice in $d$ Dimensions or not?






combinatorics geometry logic proof-writing modular-arithmetic




combinatorics geometry logic proof-writing modular-arithmetic






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 13 '18 at 19:30









Tobias KempfTobias Kempf

254




254












  • $begingroup$
    By the same argument you get that in dimension $d$, if $n$ is nice also $n+2^d-1$ is nice. Also, the only number smaller than $2^d$ that is nice is $1$. But this doesn't help a lot, I guess.
    $endgroup$
    – Servaes
    Dec 13 '18 at 19:33










  • $begingroup$
    How do you divide a square in $6$ squares?
    $endgroup$
    – ajotatxe
    Dec 13 '18 at 19:36






  • 2




    $begingroup$
    @ajotatxe: Divide a $3 times 3$ square into one $2 times 2$ and five $1 times 1$.
    $endgroup$
    – Robert Israel
    Dec 13 '18 at 19:40










  • $begingroup$
    $8$ is a bit harder: see here
    $endgroup$
    – Robert Israel
    Dec 13 '18 at 20:05






  • 1




    $begingroup$
    @RobertIsrael Wouldn't a $4times4$ divided into one $3times3$ and seven $1times1$ squares also be a dissection into $8$ squares?
    $endgroup$
    – David K
    Dec 13 '18 at 20:17




















  • $begingroup$
    By the same argument you get that in dimension $d$, if $n$ is nice also $n+2^d-1$ is nice. Also, the only number smaller than $2^d$ that is nice is $1$. But this doesn't help a lot, I guess.
    $endgroup$
    – Servaes
    Dec 13 '18 at 19:33










  • $begingroup$
    How do you divide a square in $6$ squares?
    $endgroup$
    – ajotatxe
    Dec 13 '18 at 19:36






  • 2




    $begingroup$
    @ajotatxe: Divide a $3 times 3$ square into one $2 times 2$ and five $1 times 1$.
    $endgroup$
    – Robert Israel
    Dec 13 '18 at 19:40










  • $begingroup$
    $8$ is a bit harder: see here
    $endgroup$
    – Robert Israel
    Dec 13 '18 at 20:05






  • 1




    $begingroup$
    @RobertIsrael Wouldn't a $4times4$ divided into one $3times3$ and seven $1times1$ squares also be a dissection into $8$ squares?
    $endgroup$
    – David K
    Dec 13 '18 at 20:17


















$begingroup$
By the same argument you get that in dimension $d$, if $n$ is nice also $n+2^d-1$ is nice. Also, the only number smaller than $2^d$ that is nice is $1$. But this doesn't help a lot, I guess.
$endgroup$
– Servaes
Dec 13 '18 at 19:33




$begingroup$
By the same argument you get that in dimension $d$, if $n$ is nice also $n+2^d-1$ is nice. Also, the only number smaller than $2^d$ that is nice is $1$. But this doesn't help a lot, I guess.
$endgroup$
– Servaes
Dec 13 '18 at 19:33












$begingroup$
How do you divide a square in $6$ squares?
$endgroup$
– ajotatxe
Dec 13 '18 at 19:36




$begingroup$
How do you divide a square in $6$ squares?
$endgroup$
– ajotatxe
Dec 13 '18 at 19:36




2




2




$begingroup$
@ajotatxe: Divide a $3 times 3$ square into one $2 times 2$ and five $1 times 1$.
$endgroup$
– Robert Israel
Dec 13 '18 at 19:40




$begingroup$
@ajotatxe: Divide a $3 times 3$ square into one $2 times 2$ and five $1 times 1$.
$endgroup$
– Robert Israel
Dec 13 '18 at 19:40












$begingroup$
$8$ is a bit harder: see here
$endgroup$
– Robert Israel
Dec 13 '18 at 20:05




$begingroup$
$8$ is a bit harder: see here
$endgroup$
– Robert Israel
Dec 13 '18 at 20:05




1




1




$begingroup$
@RobertIsrael Wouldn't a $4times4$ divided into one $3times3$ and seven $1times1$ squares also be a dissection into $8$ squares?
$endgroup$
– David K
Dec 13 '18 at 20:17






$begingroup$
@RobertIsrael Wouldn't a $4times4$ divided into one $3times3$ and seven $1times1$ squares also be a dissection into $8$ squares?
$endgroup$
– David K
Dec 13 '18 at 20:17












1 Answer
1






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oldest

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4












$begingroup$

For any $k$, you can divide a hypercube into $k^d$ equal hypercubes.
Thus if $n$ is nice, so is $n + k^d-1$. Now $2^d-1$ and $(2^d-1)^d-1$ are coprime, so any sufficiently large integer can be expressed as $1 + m (2^d-1) + n ((2^d-1)^d-1)$ for some $m$ and $n$, and thus is nice.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If $p$ and $q$ are coprime, any positive integer $Nge (p-1)(q-1)$ can be written as $N=ap+bq$ for non-negative integers $a,b$.
    $endgroup$
    – ajotatxe
    Dec 13 '18 at 19:43













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









4












$begingroup$

For any $k$, you can divide a hypercube into $k^d$ equal hypercubes.
Thus if $n$ is nice, so is $n + k^d-1$. Now $2^d-1$ and $(2^d-1)^d-1$ are coprime, so any sufficiently large integer can be expressed as $1 + m (2^d-1) + n ((2^d-1)^d-1)$ for some $m$ and $n$, and thus is nice.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If $p$ and $q$ are coprime, any positive integer $Nge (p-1)(q-1)$ can be written as $N=ap+bq$ for non-negative integers $a,b$.
    $endgroup$
    – ajotatxe
    Dec 13 '18 at 19:43


















4












$begingroup$

For any $k$, you can divide a hypercube into $k^d$ equal hypercubes.
Thus if $n$ is nice, so is $n + k^d-1$. Now $2^d-1$ and $(2^d-1)^d-1$ are coprime, so any sufficiently large integer can be expressed as $1 + m (2^d-1) + n ((2^d-1)^d-1)$ for some $m$ and $n$, and thus is nice.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If $p$ and $q$ are coprime, any positive integer $Nge (p-1)(q-1)$ can be written as $N=ap+bq$ for non-negative integers $a,b$.
    $endgroup$
    – ajotatxe
    Dec 13 '18 at 19:43
















4












4








4





$begingroup$

For any $k$, you can divide a hypercube into $k^d$ equal hypercubes.
Thus if $n$ is nice, so is $n + k^d-1$. Now $2^d-1$ and $(2^d-1)^d-1$ are coprime, so any sufficiently large integer can be expressed as $1 + m (2^d-1) + n ((2^d-1)^d-1)$ for some $m$ and $n$, and thus is nice.






share|cite|improve this answer









$endgroup$



For any $k$, you can divide a hypercube into $k^d$ equal hypercubes.
Thus if $n$ is nice, so is $n + k^d-1$. Now $2^d-1$ and $(2^d-1)^d-1$ are coprime, so any sufficiently large integer can be expressed as $1 + m (2^d-1) + n ((2^d-1)^d-1)$ for some $m$ and $n$, and thus is nice.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 19:38









Robert IsraelRobert Israel

325k23215469




325k23215469












  • $begingroup$
    If $p$ and $q$ are coprime, any positive integer $Nge (p-1)(q-1)$ can be written as $N=ap+bq$ for non-negative integers $a,b$.
    $endgroup$
    – ajotatxe
    Dec 13 '18 at 19:43




















  • $begingroup$
    If $p$ and $q$ are coprime, any positive integer $Nge (p-1)(q-1)$ can be written as $N=ap+bq$ for non-negative integers $a,b$.
    $endgroup$
    – ajotatxe
    Dec 13 '18 at 19:43


















$begingroup$
If $p$ and $q$ are coprime, any positive integer $Nge (p-1)(q-1)$ can be written as $N=ap+bq$ for non-negative integers $a,b$.
$endgroup$
– ajotatxe
Dec 13 '18 at 19:43






$begingroup$
If $p$ and $q$ are coprime, any positive integer $Nge (p-1)(q-1)$ can be written as $N=ap+bq$ for non-negative integers $a,b$.
$endgroup$
– ajotatxe
Dec 13 '18 at 19:43




















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