Mixing
$n!=n(n-1)(n-2)cdots 3times 2times 1 $ now let
$begingroup$
$n!=n(n-1)(n-2)cdots 3times 2times 1 $
now let
$5!m!=n! , mgeq 4$ then $n+m=?$
My Try :
case$1$ Let $n=m$:
$$5!m!=m! \ 5! neq 1$$ so $nneq m$
case$2$ Let $n>m$ :
$$5!=dfrac{n!}{m!} \ 5!=n(n-1)(n-1) . . . (n-m)$$
case$3$ Let $m>n$
$$5!m!=n! \ m(m-1)(m-2)...(m-n)=dfrac{1}{5!}=dfrac{1}{120}$$
since $dfrac{1}{120}$ is't natural number so $n>m$
Now what ?
number-theory
asked Jan 11 at 21:08
Almot1960Almot1960
2,288823
$endgroup$
add a comment |
$begingroup$
$n!=n(n-1)(n-2)cdots 3times 2times 1 $
now let
$5!m!=n! , mgeq 4$ then $n+m=?$
My Try :
case$1$ Let $n=m$:
$$5!m!=m! \ 5! neq 1$$ so $nneq m$
case$2$ Let $n>m$ :
$$5!=dfrac{n!}{m!} \ 5!=n(n-1)(n-1) . . . (n-m)$$
case$3$ Let $m>n$
$$5!m!=n! \ m(m-1)(m-2)...(m-n)=dfrac{1}{5!}=dfrac{1}{120}$$
since $dfrac{1}{120}$ is't natural number so $n>m$
Now what ?
number-theory
asked Jan 11 at 21:08
Almot1960Almot1960
2,288823
$endgroup$
$begingroup$
Isn't that $5!neq1$?
$endgroup$
– jlandercy
Jan 11 at 21:12
$begingroup$
The last term on the right hand side of case $2$ should be $(n-m+1)$. Same with the last term on the left hand side of case $3$.
$endgroup$
– pwerth
Jan 11 at 21:16
$begingroup$
It should be obvious that some of your cases do not apply. What is $5!$? Multiply it out. How many ways can you factor it, How many factorizations have consecutive integers.
$endgroup$
– Doug M
Jan 11 at 21:22
add a comment |
$begingroup$
$n!=n(n-1)(n-2)cdots 3times 2times 1 $
now let
$5!m!=n! , mgeq 4$ then $n+m=?$
My Try :
case$1$ Let $n=m$:
$$5!m!=m! \ 5! neq 1$$ so $nneq m$
case$2$ Let $n>m$ :
$$5!=dfrac{n!}{m!} \ 5!=n(n-1)(n-1) . . . (n-m)$$
case$3$ Let $m>n$
$$5!m!=n! \ m(m-1)(m-2)...(m-n)=dfrac{1}{5!}=dfrac{1}{120}$$
since $dfrac{1}{120}$ is't natural number so $n>m$
Now what ?
number-theory
asked Jan 11 at 21:08
Almot1960Almot1960
2,288823
$endgroup$
$n!=n(n-1)(n-2)cdots 3times 2times 1 $
now let
$5!m!=n! , mgeq 4$ then $n+m=?$
My Try :
case$1$ Let $n=m$:
$$5!m!=m! \ 5! neq 1$$ so $nneq m$
case$2$ Let $n>m$ :
$$5!=dfrac{n!}{m!} \ 5!=n(n-1)(n-1) . . . (n-m)$$
case$3$ Let $m>n$
$$5!m!=n! \ m(m-1)(m-2)...(m-n)=dfrac{1}{5!}=dfrac{1}{120}$$
since $dfrac{1}{120}$ is't natural number so $n>m$
Now what ?
number-theory
number-theory
asked Jan 11 at 21:08
Almot1960Almot1960
2,288823
asked Jan 11 at 21:08
Almot1960Almot1960
2,288823
edited Jan 11 at 21:18
Almot1960
asked Jan 11 at 21:08
Almot1960Almot1960
2,288823
asked Jan 11 at 21:08
Almot1960Almot1960
2,288823
asked Jan 11 at 21:08
Almot1960Almot1960
2,288823
2,288823
$begingroup$
Isn't that $5!neq1$?
$endgroup$
– jlandercy
Jan 11 at 21:12
$begingroup$
The last term on the right hand side of case $2$ should be $(n-m+1)$. Same with the last term on the left hand side of case $3$.
$endgroup$
– pwerth
Jan 11 at 21:16
$begingroup$
It should be obvious that some of your cases do not apply. What is $5!$? Multiply it out. How many ways can you factor it, How many factorizations have consecutive integers.
$endgroup$
– Doug M
Jan 11 at 21:22
add a comment |
$begingroup$
Isn't that $5!neq1$?
$endgroup$
– jlandercy
Jan 11 at 21:12
$begingroup$
The last term on the right hand side of case $2$ should be $(n-m+1)$. Same with the last term on the left hand side of case $3$.
$endgroup$
– pwerth
Jan 11 at 21:16
$begingroup$
It should be obvious that some of your cases do not apply. What is $5!$? Multiply it out. How many ways can you factor it, How many factorizations have consecutive integers.
$endgroup$
– Doug M
Jan 11 at 21:22
$begingroup$
Isn't that $5!neq1$?
$endgroup$
– jlandercy
Jan 11 at 21:12
$begingroup$
Isn't that $5!neq1$?
$endgroup$
– jlandercy
Jan 11 at 21:12
$begingroup$
The last term on the right hand side of case $2$ should be $(n-m+1)$. Same with the last term on the left hand side of case $3$.
$endgroup$
– pwerth
Jan 11 at 21:16
$begingroup$
The last term on the right hand side of case $2$ should be $(n-m+1)$. Same with the last term on the left hand side of case $3$.
$endgroup$
– pwerth
Jan 11 at 21:16
$begingroup$
It should be obvious that some of your cases do not apply. What is $5!$? Multiply it out. How many ways can you factor it, How many factorizations have consecutive integers.
$endgroup$
– Doug M
Jan 11 at 21:22
$begingroup$
It should be obvious that some of your cases do not apply. What is $5!$? Multiply it out. How many ways can you factor it, How many factorizations have consecutive integers.
$endgroup$
– Doug M
Jan 11 at 21:22
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: Obviously, $n>m$. Now, we have
$$
5! = n(n-1)cdots(m+1).
$$ In how many ways $5!=120$ can be written as a product of consecutive integer(s) starting from $m+1ge 5$?
answered Jan 11 at 21:17
SongSong
12.2k630
$endgroup$
add a comment |
$begingroup$
You can just write $5!=120=frac{n!}{m!}=n(n-1)cdots(m+1).$
Now just divide in cases: if $n-m=1$ then $n=120$, and by consequence $m=119.$
If $n-m=2$ we should write $120$ as $n(n-1),$ but this is impossible (cause $10times11=110<120$ and $11times12=132>120$).
If $n-m=3$ the solution is given by $120=6times5times4$ so that we get $n=6, m=3.$
If $n-m=4$ then $n(n-1)cdots(n-3)=120$ which implies $n=5,$ and $m=1.$
Same as the last one, if $n-m=5$ then $n(n-1)cdots(n-4)=120$ which implies $n=5,$ and $m=0.$
The difference can't be bigger because there would be too many factors to obtain $120.$
answered Jan 11 at 21:18
Riccardo CecconRiccardo Ceccon
1,060321
$endgroup$
1
$begingroup$
and $m = 1, n = 5$
$endgroup$
– Doug M
Jan 11 at 21:27
$begingroup$
Fair enough, thanks!!
$endgroup$
– Riccardo Ceccon
Jan 11 at 21:39
add a comment |
$begingroup$
This is easy to figure out by brute force. $5!=120$, and the integer $m$ must satisfy the equation $(m+1)(m+2)ldots(m+r) = 120$. However, for $r geq 3$ the integer $m$ can be no larger than 5 (because $5^3> 120$) and for $r=2$ the integer $m$ can be no larger than 11. That leaves few cases left to check.
The solution $m=119$ and $n=5!=120$ works for the equation $5!m! = n!$. So $n+m=119+120 = 239$.
As that is the only way to factor 120 into $(m+1)(m+2)ldots(m+r)$ for some positive $r$ besides 120=5! and $120 = 6times 5 times 4$ and that is the only solution for $m ge 4$.
answered Jan 11 at 21:17
MikeMike
3,951412
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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oldest
votes
$begingroup$
Hint: Obviously, $n>m$. Now, we have
$$
5! = n(n-1)cdots(m+1).
$$ In how many ways $5!=120$ can be written as a product of consecutive integer(s) starting from $m+1ge 5$?
answered Jan 11 at 21:17
SongSong
12.2k630
$endgroup$
add a comment |
$begingroup$
Hint: Obviously, $n>m$. Now, we have
$$
5! = n(n-1)cdots(m+1).
$$ In how many ways $5!=120$ can be written as a product of consecutive integer(s) starting from $m+1ge 5$?
answered Jan 11 at 21:17
SongSong
12.2k630
$endgroup$
add a comment |
$begingroup$
Hint: Obviously, $n>m$. Now, we have
$$
5! = n(n-1)cdots(m+1).
$$ In how many ways $5!=120$ can be written as a product of consecutive integer(s) starting from $m+1ge 5$?
answered Jan 11 at 21:17
SongSong
12.2k630
$endgroup$
Hint: Obviously, $n>m$. Now, we have
$$
5! = n(n-1)cdots(m+1).
$$ In how many ways $5!=120$ can be written as a product of consecutive integer(s) starting from $m+1ge 5$?
answered Jan 11 at 21:17
SongSong
12.2k630
answered Jan 11 at 21:17
SongSong
12.2k630
answered Jan 11 at 21:17
SongSong
12.2k630
answered Jan 11 at 21:17
SongSong
12.2k630
12.2k630
add a comment |
add a comment |
$begingroup$
You can just write $5!=120=frac{n!}{m!}=n(n-1)cdots(m+1).$
Now just divide in cases: if $n-m=1$ then $n=120$, and by consequence $m=119.$
If $n-m=2$ we should write $120$ as $n(n-1),$ but this is impossible (cause $10times11=110<120$ and $11times12=132>120$).
If $n-m=3$ the solution is given by $120=6times5times4$ so that we get $n=6, m=3.$
If $n-m=4$ then $n(n-1)cdots(n-3)=120$ which implies $n=5,$ and $m=1.$
Same as the last one, if $n-m=5$ then $n(n-1)cdots(n-4)=120$ which implies $n=5,$ and $m=0.$
The difference can't be bigger because there would be too many factors to obtain $120.$
answered Jan 11 at 21:18
Riccardo CecconRiccardo Ceccon
1,060321
$endgroup$
1
$begingroup$
and $m = 1, n = 5$
$endgroup$
– Doug M
Jan 11 at 21:27
$begingroup$
Fair enough, thanks!!
$endgroup$
– Riccardo Ceccon
Jan 11 at 21:39
add a comment |
$begingroup$
You can just write $5!=120=frac{n!}{m!}=n(n-1)cdots(m+1).$
Now just divide in cases: if $n-m=1$ then $n=120$, and by consequence $m=119.$
If $n-m=2$ we should write $120$ as $n(n-1),$ but this is impossible (cause $10times11=110<120$ and $11times12=132>120$).
If $n-m=3$ the solution is given by $120=6times5times4$ so that we get $n=6, m=3.$
If $n-m=4$ then $n(n-1)cdots(n-3)=120$ which implies $n=5,$ and $m=1.$
Same as the last one, if $n-m=5$ then $n(n-1)cdots(n-4)=120$ which implies $n=5,$ and $m=0.$
The difference can't be bigger because there would be too many factors to obtain $120.$
answered Jan 11 at 21:18
Riccardo CecconRiccardo Ceccon
1,060321
$endgroup$
1
$begingroup$
and $m = 1, n = 5$
$endgroup$
– Doug M
Jan 11 at 21:27
$begingroup$
Fair enough, thanks!!
$endgroup$
– Riccardo Ceccon
Jan 11 at 21:39
add a comment |
$begingroup$
You can just write $5!=120=frac{n!}{m!}=n(n-1)cdots(m+1).$
Now just divide in cases: if $n-m=1$ then $n=120$, and by consequence $m=119.$
If $n-m=2$ we should write $120$ as $n(n-1),$ but this is impossible (cause $10times11=110<120$ and $11times12=132>120$).
If $n-m=3$ the solution is given by $120=6times5times4$ so that we get $n=6, m=3.$
If $n-m=4$ then $n(n-1)cdots(n-3)=120$ which implies $n=5,$ and $m=1.$
Same as the last one, if $n-m=5$ then $n(n-1)cdots(n-4)=120$ which implies $n=5,$ and $m=0.$
The difference can't be bigger because there would be too many factors to obtain $120.$
answered Jan 11 at 21:18
Riccardo CecconRiccardo Ceccon
1,060321
$endgroup$
You can just write $5!=120=frac{n!}{m!}=n(n-1)cdots(m+1).$
Now just divide in cases: if $n-m=1$ then $n=120$, and by consequence $m=119.$
If $n-m=2$ we should write $120$ as $n(n-1),$ but this is impossible (cause $10times11=110<120$ and $11times12=132>120$).
If $n-m=3$ the solution is given by $120=6times5times4$ so that we get $n=6, m=3.$
If $n-m=4$ then $n(n-1)cdots(n-3)=120$ which implies $n=5,$ and $m=1.$
Same as the last one, if $n-m=5$ then $n(n-1)cdots(n-4)=120$ which implies $n=5,$ and $m=0.$
The difference can't be bigger because there would be too many factors to obtain $120.$
answered Jan 11 at 21:18
Riccardo CecconRiccardo Ceccon
1,060321
edited Jan 11 at 21:41
answered Jan 11 at 21:18
Riccardo CecconRiccardo Ceccon
1,060321
answered Jan 11 at 21:18
Riccardo CecconRiccardo Ceccon
1,060321
answered Jan 11 at 21:18
Riccardo CecconRiccardo Ceccon
1,060321
1,060321
1
$begingroup$
and $m = 1, n = 5$
$endgroup$
– Doug M
Jan 11 at 21:27
$begingroup$
Fair enough, thanks!!
$endgroup$
– Riccardo Ceccon
Jan 11 at 21:39
add a comment |
1
$begingroup$
and $m = 1, n = 5$
$endgroup$
– Doug M
Jan 11 at 21:27
$begingroup$
Fair enough, thanks!!
$endgroup$
– Riccardo Ceccon
Jan 11 at 21:39
1
1
$begingroup$
and $m = 1, n = 5$
$endgroup$
– Doug M
Jan 11 at 21:27
$begingroup$
and $m = 1, n = 5$
$endgroup$
– Doug M
Jan 11 at 21:27
$begingroup$
Fair enough, thanks!!
$endgroup$
– Riccardo Ceccon
Jan 11 at 21:39
$begingroup$
Fair enough, thanks!!
$endgroup$
– Riccardo Ceccon
Jan 11 at 21:39
add a comment |
$begingroup$
This is easy to figure out by brute force. $5!=120$, and the integer $m$ must satisfy the equation $(m+1)(m+2)ldots(m+r) = 120$. However, for $r geq 3$ the integer $m$ can be no larger than 5 (because $5^3> 120$) and for $r=2$ the integer $m$ can be no larger than 11. That leaves few cases left to check.
The solution $m=119$ and $n=5!=120$ works for the equation $5!m! = n!$. So $n+m=119+120 = 239$.
As that is the only way to factor 120 into $(m+1)(m+2)ldots(m+r)$ for some positive $r$ besides 120=5! and $120 = 6times 5 times 4$ and that is the only solution for $m ge 4$.
answered Jan 11 at 21:17
MikeMike
3,951412
$endgroup$
add a comment |
$begingroup$
This is easy to figure out by brute force. $5!=120$, and the integer $m$ must satisfy the equation $(m+1)(m+2)ldots(m+r) = 120$. However, for $r geq 3$ the integer $m$ can be no larger than 5 (because $5^3> 120$) and for $r=2$ the integer $m$ can be no larger than 11. That leaves few cases left to check.
The solution $m=119$ and $n=5!=120$ works for the equation $5!m! = n!$. So $n+m=119+120 = 239$.
As that is the only way to factor 120 into $(m+1)(m+2)ldots(m+r)$ for some positive $r$ besides 120=5! and $120 = 6times 5 times 4$ and that is the only solution for $m ge 4$.
answered Jan 11 at 21:17
MikeMike
3,951412
$endgroup$
add a comment |
$begingroup$
This is easy to figure out by brute force. $5!=120$, and the integer $m$ must satisfy the equation $(m+1)(m+2)ldots(m+r) = 120$. However, for $r geq 3$ the integer $m$ can be no larger than 5 (because $5^3> 120$) and for $r=2$ the integer $m$ can be no larger than 11. That leaves few cases left to check.
The solution $m=119$ and $n=5!=120$ works for the equation $5!m! = n!$. So $n+m=119+120 = 239$.
As that is the only way to factor 120 into $(m+1)(m+2)ldots(m+r)$ for some positive $r$ besides 120=5! and $120 = 6times 5 times 4$ and that is the only solution for $m ge 4$.
answered Jan 11 at 21:17
MikeMike
3,951412
$endgroup$
This is easy to figure out by brute force. $5!=120$, and the integer $m$ must satisfy the equation $(m+1)(m+2)ldots(m+r) = 120$. However, for $r geq 3$ the integer $m$ can be no larger than 5 (because $5^3> 120$) and for $r=2$ the integer $m$ can be no larger than 11. That leaves few cases left to check.
The solution $m=119$ and $n=5!=120$ works for the equation $5!m! = n!$. So $n+m=119+120 = 239$.
As that is the only way to factor 120 into $(m+1)(m+2)ldots(m+r)$ for some positive $r$ besides 120=5! and $120 = 6times 5 times 4$ and that is the only solution for $m ge 4$.
answered Jan 11 at 21:17
MikeMike
3,951412
edited Jan 11 at 21:55
answered Jan 11 at 21:17
MikeMike
3,951412
answered Jan 11 at 21:17
MikeMike
3,951412
answered Jan 11 at 21:17
MikeMike
3,951412
3,951412
add a comment |
add a comment |
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$begingroup$
Isn't that $5!neq1$?
$endgroup$
– jlandercy
Jan 11 at 21:12
$begingroup$
The last term on the right hand side of case $2$ should be $(n-m+1)$. Same with the last term on the left hand side of case $3$.
$endgroup$
– pwerth
Jan 11 at 21:16
$begingroup$
It should be obvious that some of your cases do not apply. What is $5!$? Multiply it out. How many ways can you factor it, How many factorizations have consecutive integers.
$endgroup$
– Doug M
Jan 11 at 21:22