Dividing Hypercubes into $n$ smaller Hypercubes
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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
asked Dec 13 '18 at 19:30
Tobias KempfTobias Kempf
254
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|
show 1 more comment
$begingroup$
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
asked Dec 13 '18 at 19:30
Tobias KempfTobias Kempf
254
$endgroup$
$begingroup$
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
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How do you divide a square in $6$ squares?
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– ajotatxe
Dec 13 '18 at 19:36
2
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@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
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$8$ 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
|
show 1 more comment
$begingroup$
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
asked Dec 13 '18 at 19:30
Tobias KempfTobias Kempf
254
$endgroup$
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
asked Dec 13 '18 at 19:30
Tobias KempfTobias Kempf
254
asked Dec 13 '18 at 19:30
Tobias KempfTobias Kempf
254
asked Dec 13 '18 at 19:30
Tobias KempfTobias Kempf
254
asked Dec 13 '18 at 19:30
Tobias KempfTobias Kempf
254
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 13 '18 at 19:38
Robert IsraelRobert Israel
325k23215469
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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$begingroup$
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.
answered Dec 13 '18 at 19:38
Robert IsraelRobert Israel
325k23215469
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 13 '18 at 19:38
Robert IsraelRobert Israel
325k23215469
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 13 '18 at 19:38
Robert IsraelRobert Israel
325k23215469
$endgroup$
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.
answered Dec 13 '18 at 19:38
Robert IsraelRobert Israel
325k23215469
answered Dec 13 '18 at 19:38
Robert IsraelRobert Israel
325k23215469
answered Dec 13 '18 at 19:38
Robert IsraelRobert Israel
325k23215469
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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