Pyramid Of Spheres
$begingroup$
A collection of identical spheres can be formed into a “square” pyramid (a pyramid with a base (bottom layer) made up of $n times n$ spheres whose next layer is made up of $(n-1) times (n-1)$ spheres, continuing this way up to the top layer of one sphere). The same collection of spheres can also be formed into a single-layer $k times k$ “square” where $k < 100$.
Find the largest possible value of $k$ for such a collection of spheres.
It seems that the answer will be the square of a number less than $100$ and the approach I took was saying
$$
k^2= 1+ 2^2+ ....(n-1)^2 + n^2
$$
since each "level" of the pyramid has exactly $n^2$ spheres for any given level $n$ of the pyramid. The sum of these layers or levels will be $k^2$
but how this sum can be evaluated is where Me stuck :(
combinatorics word-problem
edited Dec 3 '18 at 20:48
Daniele Tampieri
2,1621621
asked Dec 3 '18 at 20:11
RandinRandin
347116
$endgroup$
add a comment |
$begingroup$
A collection of identical spheres can be formed into a “square” pyramid (a pyramid with a base (bottom layer) made up of $n times n$ spheres whose next layer is made up of $(n-1) times (n-1)$ spheres, continuing this way up to the top layer of one sphere). The same collection of spheres can also be formed into a single-layer $k times k$ “square” where $k < 100$.
Find the largest possible value of $k$ for such a collection of spheres.
It seems that the answer will be the square of a number less than $100$ and the approach I took was saying
$$
k^2= 1+ 2^2+ ....(n-1)^2 + n^2
$$
since each "level" of the pyramid has exactly $n^2$ spheres for any given level $n$ of the pyramid. The sum of these layers or levels will be $k^2$
but how this sum can be evaluated is where Me stuck :(
combinatorics word-problem
edited Dec 3 '18 at 20:48
Daniele Tampieri
2,1621621
asked Dec 3 '18 at 20:11
RandinRandin
347116
$endgroup$
2
$begingroup$
as it happens, the only possibility is the sum of the squares up to $24$ add up to $70^2.$ You can easily find a formula for the sum of the squares from $1$ to $n^2,$ it is a cubic polynomial in $n,$ actually $frac{n(n+1)(2n+1)}{6},$ and you can prove this by induction. Setting that to $k^2$ can be done by computer up to a given bound, but the fact that this is the only answer is a matter for elliptic curves
$endgroup$
– Will Jagy
Dec 3 '18 at 20:17
$begingroup$
W. Anglin, The square pyramid puzzle, American Mathematical Monthly 97 (1990), 120-123
$endgroup$
– Will Jagy
Dec 3 '18 at 20:30
$begingroup$
will jagy there is no time to run a computer program to solve this problem on a contest ! im wondering if i got it right .. the formula can be derived how ?
$endgroup$
– Randin
Dec 3 '18 at 20:42
add a comment |
$begingroup$
A collection of identical spheres can be formed into a “square” pyramid (a pyramid with a base (bottom layer) made up of $n times n$ spheres whose next layer is made up of $(n-1) times (n-1)$ spheres, continuing this way up to the top layer of one sphere). The same collection of spheres can also be formed into a single-layer $k times k$ “square” where $k < 100$.
Find the largest possible value of $k$ for such a collection of spheres.
It seems that the answer will be the square of a number less than $100$ and the approach I took was saying
$$
k^2= 1+ 2^2+ ....(n-1)^2 + n^2
$$
since each "level" of the pyramid has exactly $n^2$ spheres for any given level $n$ of the pyramid. The sum of these layers or levels will be $k^2$
but how this sum can be evaluated is where Me stuck :(
combinatorics word-problem
edited Dec 3 '18 at 20:48
Daniele Tampieri
2,1621621
asked Dec 3 '18 at 20:11
RandinRandin
347116
$endgroup$
A collection of identical spheres can be formed into a “square” pyramid (a pyramid with a base (bottom layer) made up of $n times n$ spheres whose next layer is made up of $(n-1) times (n-1)$ spheres, continuing this way up to the top layer of one sphere). The same collection of spheres can also be formed into a single-layer $k times k$ “square” where $k < 100$.
Find the largest possible value of $k$ for such a collection of spheres.
It seems that the answer will be the square of a number less than $100$ and the approach I took was saying
$$
k^2= 1+ 2^2+ ....(n-1)^2 + n^2
$$
since each "level" of the pyramid has exactly $n^2$ spheres for any given level $n$ of the pyramid. The sum of these layers or levels will be $k^2$
but how this sum can be evaluated is where Me stuck :(
combinatorics word-problem
combinatorics word-problem
edited Dec 3 '18 at 20:48
Daniele Tampieri
2,1621621
asked Dec 3 '18 at 20:11
RandinRandin
347116
edited Dec 3 '18 at 20:48
Daniele Tampieri
2,1621621
asked Dec 3 '18 at 20:11
RandinRandin
347116
edited Dec 3 '18 at 20:48
Daniele Tampieri
2,1621621
edited Dec 3 '18 at 20:48
Daniele Tampieri
2,1621621
edited Dec 3 '18 at 20:48
Daniele Tampieri
2,1621621
2,1621621
asked Dec 3 '18 at 20:11
RandinRandin
347116
asked Dec 3 '18 at 20:11
RandinRandin
347116
asked Dec 3 '18 at 20:11
RandinRandin
347116
347116
2
$begingroup$
as it happens, the only possibility is the sum of the squares up to $24$ add up to $70^2.$ You can easily find a formula for the sum of the squares from $1$ to $n^2,$ it is a cubic polynomial in $n,$ actually $frac{n(n+1)(2n+1)}{6},$ and you can prove this by induction. Setting that to $k^2$ can be done by computer up to a given bound, but the fact that this is the only answer is a matter for elliptic curves
$endgroup$
– Will Jagy
Dec 3 '18 at 20:17
$begingroup$
W. Anglin, The square pyramid puzzle, American Mathematical Monthly 97 (1990), 120-123
$endgroup$
– Will Jagy
Dec 3 '18 at 20:30
$begingroup$
will jagy there is no time to run a computer program to solve this problem on a contest ! im wondering if i got it right .. the formula can be derived how ?
$endgroup$
– Randin
Dec 3 '18 at 20:42
add a comment |
2
$begingroup$
as it happens, the only possibility is the sum of the squares up to $24$ add up to $70^2.$ You can easily find a formula for the sum of the squares from $1$ to $n^2,$ it is a cubic polynomial in $n,$ actually $frac{n(n+1)(2n+1)}{6},$ and you can prove this by induction. Setting that to $k^2$ can be done by computer up to a given bound, but the fact that this is the only answer is a matter for elliptic curves
$endgroup$
– Will Jagy
Dec 3 '18 at 20:17
$begingroup$
W. Anglin, The square pyramid puzzle, American Mathematical Monthly 97 (1990), 120-123
$endgroup$
– Will Jagy
Dec 3 '18 at 20:30
$begingroup$
will jagy there is no time to run a computer program to solve this problem on a contest ! im wondering if i got it right .. the formula can be derived how ?
$endgroup$
– Randin
Dec 3 '18 at 20:42
2
2
$begingroup$
as it happens, the only possibility is the sum of the squares up to $24$ add up to $70^2.$ You can easily find a formula for the sum of the squares from $1$ to $n^2,$ it is a cubic polynomial in $n,$ actually $frac{n(n+1)(2n+1)}{6},$ and you can prove this by induction. Setting that to $k^2$ can be done by computer up to a given bound, but the fact that this is the only answer is a matter for elliptic curves
$endgroup$
– Will Jagy
Dec 3 '18 at 20:17
$begingroup$
as it happens, the only possibility is the sum of the squares up to $24$ add up to $70^2.$ You can easily find a formula for the sum of the squares from $1$ to $n^2,$ it is a cubic polynomial in $n,$ actually $frac{n(n+1)(2n+1)}{6},$ and you can prove this by induction. Setting that to $k^2$ can be done by computer up to a given bound, but the fact that this is the only answer is a matter for elliptic curves
$endgroup$
– Will Jagy
Dec 3 '18 at 20:17
$begingroup$
W. Anglin, The square pyramid puzzle, American Mathematical Monthly 97 (1990), 120-123
$endgroup$
– Will Jagy
Dec 3 '18 at 20:30
$begingroup$
W. Anglin, The square pyramid puzzle, American Mathematical Monthly 97 (1990), 120-123
$endgroup$
– Will Jagy
Dec 3 '18 at 20:30
$begingroup$
will jagy there is no time to run a computer program to solve this problem on a contest ! im wondering if i got it right .. the formula can be derived how ?
$endgroup$
– Randin
Dec 3 '18 at 20:42
$begingroup$
will jagy there is no time to run a computer program to solve this problem on a contest ! im wondering if i got it right .. the formula can be derived how ?
$endgroup$
– Randin
Dec 3 '18 at 20:42
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
If we impose the bound $1<k<100$ (the problem implies more than one ball), then uniqueness can be solved easily without extensive computational effort or the full use of elliptic curves.
Start with the $n(n+1)(2n+1)/6$ formula for the sum of the first $n$ positive squares. We must have this equal to a square $k^2$ itself, with $k<100$ so we infer $n^3<60,000$, thus $1<n<40$.
Then $n(n+1)(2n+1)=6k^2$ where all the factors on the left side are pairwise relatively prime. Therefore $n$ must be $in{1,2,3,6}$ times a perfect square, similarly for $n+1$ and $2n+1$.
We therefore rule out any cases where any of these factors is $equiv 5bmod 6$ as such a number cannot satisfy the constraint described above. This forces $nin {0,1,3}bmod 6$.
If $n$ is multiple of $3$ then $2n+1$ must be odd, not a multiple of $3$ and hence a perfect square, and with $1<n<40$ this forces $nin{12,24}$. If $nequiv 1bmod 6$ then it must be a perfect square, and only $25$ among these possibilities is within the bounds $1<n<40$. Then the only remaining possible values of $n$ which could give a sum of $k^2$ with $1<k<100$ are $12,24,25$. The first and last of these give an unsquared factor of $13$ when substituted into $n(n+1)(2n+1)/6$, leaving $n=24,k=70$ as the sole survivor.
answered Dec 3 '18 at 21:30
Oscar LanziOscar Lanzi
12.4k12036
$endgroup$
add a comment |
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$begingroup$
If we impose the bound $1<k<100$ (the problem implies more than one ball), then uniqueness can be solved easily without extensive computational effort or the full use of elliptic curves.
Start with the $n(n+1)(2n+1)/6$ formula for the sum of the first $n$ positive squares. We must have this equal to a square $k^2$ itself, with $k<100$ so we infer $n^3<60,000$, thus $1<n<40$.
Then $n(n+1)(2n+1)=6k^2$ where all the factors on the left side are pairwise relatively prime. Therefore $n$ must be $in{1,2,3,6}$ times a perfect square, similarly for $n+1$ and $2n+1$.
We therefore rule out any cases where any of these factors is $equiv 5bmod 6$ as such a number cannot satisfy the constraint described above. This forces $nin {0,1,3}bmod 6$.
If $n$ is multiple of $3$ then $2n+1$ must be odd, not a multiple of $3$ and hence a perfect square, and with $1<n<40$ this forces $nin{12,24}$. If $nequiv 1bmod 6$ then it must be a perfect square, and only $25$ among these possibilities is within the bounds $1<n<40$. Then the only remaining possible values of $n$ which could give a sum of $k^2$ with $1<k<100$ are $12,24,25$. The first and last of these give an unsquared factor of $13$ when substituted into $n(n+1)(2n+1)/6$, leaving $n=24,k=70$ as the sole survivor.
answered Dec 3 '18 at 21:30
Oscar LanziOscar Lanzi
12.4k12036
$endgroup$
add a comment |
$begingroup$
If we impose the bound $1<k<100$ (the problem implies more than one ball), then uniqueness can be solved easily without extensive computational effort or the full use of elliptic curves.
Start with the $n(n+1)(2n+1)/6$ formula for the sum of the first $n$ positive squares. We must have this equal to a square $k^2$ itself, with $k<100$ so we infer $n^3<60,000$, thus $1<n<40$.
Then $n(n+1)(2n+1)=6k^2$ where all the factors on the left side are pairwise relatively prime. Therefore $n$ must be $in{1,2,3,6}$ times a perfect square, similarly for $n+1$ and $2n+1$.
We therefore rule out any cases where any of these factors is $equiv 5bmod 6$ as such a number cannot satisfy the constraint described above. This forces $nin {0,1,3}bmod 6$.
If $n$ is multiple of $3$ then $2n+1$ must be odd, not a multiple of $3$ and hence a perfect square, and with $1<n<40$ this forces $nin{12,24}$. If $nequiv 1bmod 6$ then it must be a perfect square, and only $25$ among these possibilities is within the bounds $1<n<40$. Then the only remaining possible values of $n$ which could give a sum of $k^2$ with $1<k<100$ are $12,24,25$. The first and last of these give an unsquared factor of $13$ when substituted into $n(n+1)(2n+1)/6$, leaving $n=24,k=70$ as the sole survivor.
answered Dec 3 '18 at 21:30
Oscar LanziOscar Lanzi
12.4k12036
$endgroup$
add a comment |
$begingroup$
If we impose the bound $1<k<100$ (the problem implies more than one ball), then uniqueness can be solved easily without extensive computational effort or the full use of elliptic curves.
Start with the $n(n+1)(2n+1)/6$ formula for the sum of the first $n$ positive squares. We must have this equal to a square $k^2$ itself, with $k<100$ so we infer $n^3<60,000$, thus $1<n<40$.
Then $n(n+1)(2n+1)=6k^2$ where all the factors on the left side are pairwise relatively prime. Therefore $n$ must be $in{1,2,3,6}$ times a perfect square, similarly for $n+1$ and $2n+1$.
We therefore rule out any cases where any of these factors is $equiv 5bmod 6$ as such a number cannot satisfy the constraint described above. This forces $nin {0,1,3}bmod 6$.
If $n$ is multiple of $3$ then $2n+1$ must be odd, not a multiple of $3$ and hence a perfect square, and with $1<n<40$ this forces $nin{12,24}$. If $nequiv 1bmod 6$ then it must be a perfect square, and only $25$ among these possibilities is within the bounds $1<n<40$. Then the only remaining possible values of $n$ which could give a sum of $k^2$ with $1<k<100$ are $12,24,25$. The first and last of these give an unsquared factor of $13$ when substituted into $n(n+1)(2n+1)/6$, leaving $n=24,k=70$ as the sole survivor.
answered Dec 3 '18 at 21:30
Oscar LanziOscar Lanzi
12.4k12036
$endgroup$
If we impose the bound $1<k<100$ (the problem implies more than one ball), then uniqueness can be solved easily without extensive computational effort or the full use of elliptic curves.
Start with the $n(n+1)(2n+1)/6$ formula for the sum of the first $n$ positive squares. We must have this equal to a square $k^2$ itself, with $k<100$ so we infer $n^3<60,000$, thus $1<n<40$.
Then $n(n+1)(2n+1)=6k^2$ where all the factors on the left side are pairwise relatively prime. Therefore $n$ must be $in{1,2,3,6}$ times a perfect square, similarly for $n+1$ and $2n+1$.
We therefore rule out any cases where any of these factors is $equiv 5bmod 6$ as such a number cannot satisfy the constraint described above. This forces $nin {0,1,3}bmod 6$.
If $n$ is multiple of $3$ then $2n+1$ must be odd, not a multiple of $3$ and hence a perfect square, and with $1<n<40$ this forces $nin{12,24}$. If $nequiv 1bmod 6$ then it must be a perfect square, and only $25$ among these possibilities is within the bounds $1<n<40$. Then the only remaining possible values of $n$ which could give a sum of $k^2$ with $1<k<100$ are $12,24,25$. The first and last of these give an unsquared factor of $13$ when substituted into $n(n+1)(2n+1)/6$, leaving $n=24,k=70$ as the sole survivor.
answered Dec 3 '18 at 21:30
Oscar LanziOscar Lanzi
12.4k12036
edited Dec 3 '18 at 21:46
answered Dec 3 '18 at 21:30
Oscar LanziOscar Lanzi
12.4k12036
answered Dec 3 '18 at 21:30
Oscar LanziOscar Lanzi
12.4k12036
answered Dec 3 '18 at 21:30
Oscar LanziOscar Lanzi
12.4k12036
12.4k12036
add a comment |
add a comment |
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$begingroup$
as it happens, the only possibility is the sum of the squares up to $24$ add up to $70^2.$ You can easily find a formula for the sum of the squares from $1$ to $n^2,$ it is a cubic polynomial in $n,$ actually $frac{n(n+1)(2n+1)}{6},$ and you can prove this by induction. Setting that to $k^2$ can be done by computer up to a given bound, but the fact that this is the only answer is a matter for elliptic curves
$endgroup$
– Will Jagy
Dec 3 '18 at 20:17
$begingroup$
W. Anglin, The square pyramid puzzle, American Mathematical Monthly 97 (1990), 120-123
$endgroup$
– Will Jagy
Dec 3 '18 at 20:30
$begingroup$
will jagy there is no time to run a computer program to solve this problem on a contest ! im wondering if i got it right .. the formula can be derived how ?
$endgroup$
– Randin
Dec 3 '18 at 20:42