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Solving algebraic equations involving factorials - without trial and error.
Clash Royale CLAN TAG#URR8PPP
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0
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The question is defined as follows:
$$frac(x!)^3x-1=3455$$
I first did the basics which was getting rid of the 1 onto the left and getting rid of the $x$, then I factorised both sides to get an expression
as seen below:
$$x!=3456^1/3cdot x^1/3$$
$$x(x-1)!=xcdot 3456^1/3cdot x^-2/3$$
$$3456^1/3/(x-1)!=x^2/3$$
I then got rid of the power of x and tried using an iterative process, however it got messy because negative integers and square roots don’t go well together.
So I wanted to know: can you solve this question without trial and error? And if so, please can you exemplify your justification in mathematical notation?
algebraic-equations
edited Jul 24 at 17:37
NickD
9181412
asked Jul 24 at 17:11
Benny Copeland
1
add a comment |Â
up vote
0
down vote
favorite
The question is defined as follows:
$$frac(x!)^3x-1=3455$$
I first did the basics which was getting rid of the 1 onto the left and getting rid of the $x$, then I factorised both sides to get an expression
as seen below:
$$x!=3456^1/3cdot x^1/3$$
$$x(x-1)!=xcdot 3456^1/3cdot x^-2/3$$
$$3456^1/3/(x-1)!=x^2/3$$
I then got rid of the power of x and tried using an iterative process, however it got messy because negative integers and square roots don’t go well together.
So I wanted to know: can you solve this question without trial and error? And if so, please can you exemplify your justification in mathematical notation?
algebraic-equations
edited Jul 24 at 17:37
NickD
9181412
asked Jul 24 at 17:11
Benny Copeland
1
2
I'd get the prime factorization of 3456 and then try matching powers of the prime factors.
– NickD
Jul 24 at 17:18
Please use MathJax the next time you are asking an question. I just fixed that for you.
– mrtaurho
Jul 24 at 17:18
Maybe you were only supposed to find the small "nice" positive root, which can be guessed fairly easily.
– dxiv
Jul 24 at 17:29
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The question is defined as follows:
$$frac(x!)^3x-1=3455$$
I first did the basics which was getting rid of the 1 onto the left and getting rid of the $x$, then I factorised both sides to get an expression
as seen below:
$$x!=3456^1/3cdot x^1/3$$
$$x(x-1)!=xcdot 3456^1/3cdot x^-2/3$$
$$3456^1/3/(x-1)!=x^2/3$$
I then got rid of the power of x and tried using an iterative process, however it got messy because negative integers and square roots don’t go well together.
So I wanted to know: can you solve this question without trial and error? And if so, please can you exemplify your justification in mathematical notation?
algebraic-equations
edited Jul 24 at 17:37
NickD
9181412
asked Jul 24 at 17:11
Benny Copeland
1
The question is defined as follows:
$$frac(x!)^3x-1=3455$$
I first did the basics which was getting rid of the 1 onto the left and getting rid of the $x$, then I factorised both sides to get an expression
as seen below:
$$x!=3456^1/3cdot x^1/3$$
$$x(x-1)!=xcdot 3456^1/3cdot x^-2/3$$
$$3456^1/3/(x-1)!=x^2/3$$
I then got rid of the power of x and tried using an iterative process, however it got messy because negative integers and square roots don’t go well together.
So I wanted to know: can you solve this question without trial and error? And if so, please can you exemplify your justification in mathematical notation?
algebraic-equations
edited Jul 24 at 17:37
NickD
9181412
asked Jul 24 at 17:11
Benny Copeland
1
edited Jul 24 at 17:37
NickD
9181412
edited Jul 24 at 17:37
NickD
9181412
edited Jul 24 at 17:37
NickD
9181412
9181412
asked Jul 24 at 17:11
Benny Copeland
1
asked Jul 24 at 17:11
Benny Copeland
1
asked Jul 24 at 17:11
Benny Copeland
1
1
2
I'd get the prime factorization of 3456 and then try matching powers of the prime factors.
– NickD
Jul 24 at 17:18
Please use MathJax the next time you are asking an question. I just fixed that for you.
– mrtaurho
Jul 24 at 17:18
Maybe you were only supposed to find the small "nice" positive root, which can be guessed fairly easily.
– dxiv
Jul 24 at 17:29
add a comment |Â
2
I'd get the prime factorization of 3456 and then try matching powers of the prime factors.
– NickD
Jul 24 at 17:18
Please use MathJax the next time you are asking an question. I just fixed that for you.
– mrtaurho
Jul 24 at 17:18
Maybe you were only supposed to find the small "nice" positive root, which can be guessed fairly easily.
– dxiv
Jul 24 at 17:29
2
2
I'd get the prime factorization of 3456 and then try matching powers of the prime factors.
– NickD
Jul 24 at 17:18
I'd get the prime factorization of 3456 and then try matching powers of the prime factors.
– NickD
Jul 24 at 17:18
Please use MathJax the next time you are asking an question. I just fixed that for you.
– mrtaurho
Jul 24 at 17:18
Please use MathJax the next time you are asking an question. I just fixed that for you.
– mrtaurho
Jul 24 at 17:18
Maybe you were only supposed to find the small "nice" positive root, which can be guessed fairly easily.
– dxiv
Jul 24 at 17:29
Maybe you were only supposed to find the small "nice" positive root, which can be guessed fairly easily.
– dxiv
Jul 24 at 17:29
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
We have $3456=2^7cdot 3^3$. So, you have
$$(x!)^3=3456x=2^7cdot 3^3cdot x.$$
In particular, no prime $>3$ can divide $x!$, and so $xleq 4$. You also need $2^7 cdot 3^3 cdot x$ to be a cube, so $2x$ should be a cube, which means $x=4$. Now you can just test it and it works! so you're done.
answered Jul 24 at 17:29
Carl Schildkraut
8,26211238
Ok I’ve tested and it works thanks
– Benny Copeland
Jul 24 at 17:30
add a comment |Â
up vote
1
down vote
"Without trial and error" seems excessive, as the function is very quickly growing.
Grossly, you can use
$$x!^2le fracx!^3xle x!^3$$ which gives you the range
$$x!in(15,59)$$
and only $4!$ can do.
But my first reaction would be to try $3$ to $5$ without any preliminary effort.
It is worth to notice that Stirling is not helpful here.
For larger values, I would consider the bracketing between $fraclog(m+1)3loglog(m+1)$ and $fraclog(m+1)2loglog(m+1)$ and work by dichotomy.
answered Jul 24 at 17:59
Yves Daoust
111k665203
add a comment |Â
up vote
-1
down vote
Let $,a_n=dfracleft(n!right)^3n,$ then $,a_n+1=n(n+1)^2a_n,$. It's quite obvious that $,a_n,$ grows very fast, and the first few terms are $,a_1=1, a_2=4, a_3 = 72, a_4=3456,$. The latter gives the solution $,x=4,$.
answered Jul 24 at 18:08
dxiv
53.9k64796
@YvesDaoust It starts as a (computationally cheaper) way to narrow down $,n = lfloor x rfloor,$, which would be useful in any approach to solve the problem. Then, yes, it so happens that it does in fact find the solution. FWIW all of the answers posted thus far are one form or another of trial and error.
– dxiv
Jul 24 at 18:25
@YvesDaoust Fine, I can restate the conclusion in my answer as $,x in [4, 5),$ instead ;-) It's still finding the range with less calculations than factoring or extracting roots.
– dxiv
Jul 24 at 18:46
@YvesDaoust I'll just note that the recursive calculation scales better, and leave it at that. As an example, the other two methods don't work well if the RHS were $6220799$ instead of $3455$.
– dxiv
Jul 24 at 19:05
You probably mean $62207999$.
– Yves Daoust
Jul 25 at 6:22
@YvesDaoust Right, of course. And thank you for the downvote, though it wouldn't be appropriate to return the favor.
– dxiv
Jul 25 at 6:34
 |Â
show 5 more comments
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
We have $3456=2^7cdot 3^3$. So, you have
$$(x!)^3=3456x=2^7cdot 3^3cdot x.$$
In particular, no prime $>3$ can divide $x!$, and so $xleq 4$. You also need $2^7 cdot 3^3 cdot x$ to be a cube, so $2x$ should be a cube, which means $x=4$. Now you can just test it and it works! so you're done.
answered Jul 24 at 17:29
Carl Schildkraut
8,26211238
Ok I’ve tested and it works thanks
– Benny Copeland
Jul 24 at 17:30
add a comment |Â
up vote
1
down vote
We have $3456=2^7cdot 3^3$. So, you have
$$(x!)^3=3456x=2^7cdot 3^3cdot x.$$
In particular, no prime $>3$ can divide $x!$, and so $xleq 4$. You also need $2^7 cdot 3^3 cdot x$ to be a cube, so $2x$ should be a cube, which means $x=4$. Now you can just test it and it works! so you're done.
answered Jul 24 at 17:29
Carl Schildkraut
8,26211238
Ok I’ve tested and it works thanks
– Benny Copeland
Jul 24 at 17:30
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We have $3456=2^7cdot 3^3$. So, you have
$$(x!)^3=3456x=2^7cdot 3^3cdot x.$$
In particular, no prime $>3$ can divide $x!$, and so $xleq 4$. You also need $2^7 cdot 3^3 cdot x$ to be a cube, so $2x$ should be a cube, which means $x=4$. Now you can just test it and it works! so you're done.
answered Jul 24 at 17:29
Carl Schildkraut
8,26211238
We have $3456=2^7cdot 3^3$. So, you have
$$(x!)^3=3456x=2^7cdot 3^3cdot x.$$
In particular, no prime $>3$ can divide $x!$, and so $xleq 4$. You also need $2^7 cdot 3^3 cdot x$ to be a cube, so $2x$ should be a cube, which means $x=4$. Now you can just test it and it works! so you're done.
answered Jul 24 at 17:29
Carl Schildkraut
8,26211238
answered Jul 24 at 17:29
Carl Schildkraut
8,26211238
answered Jul 24 at 17:29
Carl Schildkraut
8,26211238
answered Jul 24 at 17:29
Carl Schildkraut
8,26211238
8,26211238
Ok I’ve tested and it works thanks
– Benny Copeland
Jul 24 at 17:30
add a comment |Â
Ok I’ve tested and it works thanks
– Benny Copeland
Jul 24 at 17:30
Ok I’ve tested and it works thanks
– Benny Copeland
Jul 24 at 17:30
Ok I’ve tested and it works thanks
– Benny Copeland
Jul 24 at 17:30
add a comment |Â
up vote
1
down vote
"Without trial and error" seems excessive, as the function is very quickly growing.
Grossly, you can use
$$x!^2le fracx!^3xle x!^3$$ which gives you the range
$$x!in(15,59)$$
and only $4!$ can do.
But my first reaction would be to try $3$ to $5$ without any preliminary effort.
It is worth to notice that Stirling is not helpful here.
For larger values, I would consider the bracketing between $fraclog(m+1)3loglog(m+1)$ and $fraclog(m+1)2loglog(m+1)$ and work by dichotomy.
answered Jul 24 at 17:59
Yves Daoust
111k665203
add a comment |Â
up vote
1
down vote
"Without trial and error" seems excessive, as the function is very quickly growing.
Grossly, you can use
$$x!^2le fracx!^3xle x!^3$$ which gives you the range
$$x!in(15,59)$$
and only $4!$ can do.
But my first reaction would be to try $3$ to $5$ without any preliminary effort.
It is worth to notice that Stirling is not helpful here.
For larger values, I would consider the bracketing between $fraclog(m+1)3loglog(m+1)$ and $fraclog(m+1)2loglog(m+1)$ and work by dichotomy.
answered Jul 24 at 17:59
Yves Daoust
111k665203
add a comment |Â
up vote
1
down vote
up vote
1
down vote
"Without trial and error" seems excessive, as the function is very quickly growing.
Grossly, you can use
$$x!^2le fracx!^3xle x!^3$$ which gives you the range
$$x!in(15,59)$$
and only $4!$ can do.
But my first reaction would be to try $3$ to $5$ without any preliminary effort.
It is worth to notice that Stirling is not helpful here.
For larger values, I would consider the bracketing between $fraclog(m+1)3loglog(m+1)$ and $fraclog(m+1)2loglog(m+1)$ and work by dichotomy.
answered Jul 24 at 17:59
Yves Daoust
111k665203
"Without trial and error" seems excessive, as the function is very quickly growing.
Grossly, you can use
$$x!^2le fracx!^3xle x!^3$$ which gives you the range
$$x!in(15,59)$$
and only $4!$ can do.
But my first reaction would be to try $3$ to $5$ without any preliminary effort.
It is worth to notice that Stirling is not helpful here.
For larger values, I would consider the bracketing between $fraclog(m+1)3loglog(m+1)$ and $fraclog(m+1)2loglog(m+1)$ and work by dichotomy.
answered Jul 24 at 17:59
Yves Daoust
111k665203
edited Jul 25 at 7:01
answered Jul 24 at 17:59
Yves Daoust
111k665203
answered Jul 24 at 17:59
Yves Daoust
111k665203
answered Jul 24 at 17:59
Yves Daoust
111k665203
111k665203
add a comment |Â
add a comment |Â
up vote
-1
down vote
Let $,a_n=dfracleft(n!right)^3n,$ then $,a_n+1=n(n+1)^2a_n,$. It's quite obvious that $,a_n,$ grows very fast, and the first few terms are $,a_1=1, a_2=4, a_3 = 72, a_4=3456,$. The latter gives the solution $,x=4,$.
answered Jul 24 at 18:08
dxiv
53.9k64796
@YvesDaoust It starts as a (computationally cheaper) way to narrow down $,n = lfloor x rfloor,$, which would be useful in any approach to solve the problem. Then, yes, it so happens that it does in fact find the solution. FWIW all of the answers posted thus far are one form or another of trial and error.
– dxiv
Jul 24 at 18:25
@YvesDaoust Fine, I can restate the conclusion in my answer as $,x in [4, 5),$ instead ;-) It's still finding the range with less calculations than factoring or extracting roots.
– dxiv
Jul 24 at 18:46
@YvesDaoust I'll just note that the recursive calculation scales better, and leave it at that. As an example, the other two methods don't work well if the RHS were $6220799$ instead of $3455$.
– dxiv
Jul 24 at 19:05
You probably mean $62207999$.
– Yves Daoust
Jul 25 at 6:22
@YvesDaoust Right, of course. And thank you for the downvote, though it wouldn't be appropriate to return the favor.
– dxiv
Jul 25 at 6:34
 |Â
show 5 more comments
up vote
-1
down vote
Let $,a_n=dfracleft(n!right)^3n,$ then $,a_n+1=n(n+1)^2a_n,$. It's quite obvious that $,a_n,$ grows very fast, and the first few terms are $,a_1=1, a_2=4, a_3 = 72, a_4=3456,$. The latter gives the solution $,x=4,$.
answered Jul 24 at 18:08
dxiv
53.9k64796
@YvesDaoust It starts as a (computationally cheaper) way to narrow down $,n = lfloor x rfloor,$, which would be useful in any approach to solve the problem. Then, yes, it so happens that it does in fact find the solution. FWIW all of the answers posted thus far are one form or another of trial and error.
– dxiv
Jul 24 at 18:25
@YvesDaoust Fine, I can restate the conclusion in my answer as $,x in [4, 5),$ instead ;-) It's still finding the range with less calculations than factoring or extracting roots.
– dxiv
Jul 24 at 18:46
@YvesDaoust I'll just note that the recursive calculation scales better, and leave it at that. As an example, the other two methods don't work well if the RHS were $6220799$ instead of $3455$.
– dxiv
Jul 24 at 19:05
You probably mean $62207999$.
– Yves Daoust
Jul 25 at 6:22
@YvesDaoust Right, of course. And thank you for the downvote, though it wouldn't be appropriate to return the favor.
– dxiv
Jul 25 at 6:34
 |Â
show 5 more comments
up vote
-1
down vote
up vote
-1
down vote
Let $,a_n=dfracleft(n!right)^3n,$ then $,a_n+1=n(n+1)^2a_n,$. It's quite obvious that $,a_n,$ grows very fast, and the first few terms are $,a_1=1, a_2=4, a_3 = 72, a_4=3456,$. The latter gives the solution $,x=4,$.
answered Jul 24 at 18:08
dxiv
53.9k64796
Let $,a_n=dfracleft(n!right)^3n,$ then $,a_n+1=n(n+1)^2a_n,$. It's quite obvious that $,a_n,$ grows very fast, and the first few terms are $,a_1=1, a_2=4, a_3 = 72, a_4=3456,$. The latter gives the solution $,x=4,$.
answered Jul 24 at 18:08
dxiv
53.9k64796
answered Jul 24 at 18:08
dxiv
53.9k64796
answered Jul 24 at 18:08
dxiv
53.9k64796
answered Jul 24 at 18:08
dxiv
53.9k64796
53.9k64796
@YvesDaoust It starts as a (computationally cheaper) way to narrow down $,n = lfloor x rfloor,$, which would be useful in any approach to solve the problem. Then, yes, it so happens that it does in fact find the solution. FWIW all of the answers posted thus far are one form or another of trial and error.
– dxiv
Jul 24 at 18:25
@YvesDaoust Fine, I can restate the conclusion in my answer as $,x in [4, 5),$ instead ;-) It's still finding the range with less calculations than factoring or extracting roots.
– dxiv
Jul 24 at 18:46
@YvesDaoust I'll just note that the recursive calculation scales better, and leave it at that. As an example, the other two methods don't work well if the RHS were $6220799$ instead of $3455$.
– dxiv
Jul 24 at 19:05
You probably mean $62207999$.
– Yves Daoust
Jul 25 at 6:22
@YvesDaoust Right, of course. And thank you for the downvote, though it wouldn't be appropriate to return the favor.
– dxiv
Jul 25 at 6:34
 |Â
show 5 more comments
@YvesDaoust It starts as a (computationally cheaper) way to narrow down $,n = lfloor x rfloor,$, which would be useful in any approach to solve the problem. Then, yes, it so happens that it does in fact find the solution. FWIW all of the answers posted thus far are one form or another of trial and error.
– dxiv
Jul 24 at 18:25
@YvesDaoust Fine, I can restate the conclusion in my answer as $,x in [4, 5),$ instead ;-) It's still finding the range with less calculations than factoring or extracting roots.
– dxiv
Jul 24 at 18:46
@YvesDaoust I'll just note that the recursive calculation scales better, and leave it at that. As an example, the other two methods don't work well if the RHS were $6220799$ instead of $3455$.
– dxiv
Jul 24 at 19:05
You probably mean $62207999$.
– Yves Daoust
Jul 25 at 6:22
@YvesDaoust Right, of course. And thank you for the downvote, though it wouldn't be appropriate to return the favor.
– dxiv
Jul 25 at 6:34
@YvesDaoust It starts as a (computationally cheaper) way to narrow down $,n = lfloor x rfloor,$, which would be useful in any approach to solve the problem. Then, yes, it so happens that it does in fact find the solution. FWIW all of the answers posted thus far are one form or another of trial and error.
– dxiv
Jul 24 at 18:25
@YvesDaoust It starts as a (computationally cheaper) way to narrow down $,n = lfloor x rfloor,$, which would be useful in any approach to solve the problem. Then, yes, it so happens that it does in fact find the solution. FWIW all of the answers posted thus far are one form or another of trial and error.
– dxiv
Jul 24 at 18:25
@YvesDaoust Fine, I can restate the conclusion in my answer as $,x in [4, 5),$ instead ;-) It's still finding the range with less calculations than factoring or extracting roots.
– dxiv
Jul 24 at 18:46
@YvesDaoust Fine, I can restate the conclusion in my answer as $,x in [4, 5),$ instead ;-) It's still finding the range with less calculations than factoring or extracting roots.
– dxiv
Jul 24 at 18:46
@YvesDaoust I'll just note that the recursive calculation scales better, and leave it at that. As an example, the other two methods don't work well if the RHS were $6220799$ instead of $3455$.
– dxiv
Jul 24 at 19:05
@YvesDaoust I'll just note that the recursive calculation scales better, and leave it at that. As an example, the other two methods don't work well if the RHS were $6220799$ instead of $3455$.
– dxiv
Jul 24 at 19:05
You probably mean $62207999$.
– Yves Daoust
Jul 25 at 6:22
You probably mean $62207999$.
– Yves Daoust
Jul 25 at 6:22
@YvesDaoust Right, of course. And thank you for the downvote, though it wouldn't be appropriate to return the favor.
– dxiv
Jul 25 at 6:34
@YvesDaoust Right, of course. And thank you for the downvote, though it wouldn't be appropriate to return the favor.
– dxiv
Jul 25 at 6:34
 |Â
show 5 more comments
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2
I'd get the prime factorization of 3456 and then try matching powers of the prime factors.
– NickD
Jul 24 at 17:18
Please use MathJax the next time you are asking an question. I just fixed that for you.
– mrtaurho
Jul 24 at 17:18
Maybe you were only supposed to find the small "nice" positive root, which can be guessed fairly easily.
– dxiv
Jul 24 at 17:29