How to complete the whole without fractions [on hold]
Let's say I need to split the following whole number 4 ways: 1659. I cannot have any remainders, yet I need the sum of the four divisors to equal the sum of the dividend. Is there any mathematical formula that solves this?
division-ring
edited Jan 4 at 2:11
David G. Stork
9,96021232
asked Jan 4 at 1:51
Eric HondzinskiEric Hondzinski
11
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Eric Hondzinski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as unclear what you're asking by Hans Lundmark, user91500, Paul Frost, Cesareo, mrtaurho Jan 4 at 13:34
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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Let's say I need to split the following whole number 4 ways: 1659. I cannot have any remainders, yet I need the sum of the four divisors to equal the sum of the dividend. Is there any mathematical formula that solves this?
division-ring
edited Jan 4 at 2:11
David G. Stork
9,96021232
asked Jan 4 at 1:51
Eric HondzinskiEric Hondzinski
11
New contributor
Eric Hondzinski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as unclear what you're asking by Hans Lundmark, user91500, Paul Frost, Cesareo, mrtaurho Jan 4 at 13:34
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
if we add the digits we get $1+6+5+9 = 21,$ suggesting that 3 is a factor. $1659=3cdot 553$ Now look for factors of $553$ next smallest number that might be a factor is 7.
– Doug M
Jan 4 at 1:56
3
Eric, I have no idea what you are talking about. Is there some number other than 1659 for which you can solve your question completely, and edit that information into the question?
– Will Jagy
Jan 4 at 2:03
add a comment |
Let's say I need to split the following whole number 4 ways: 1659. I cannot have any remainders, yet I need the sum of the four divisors to equal the sum of the dividend. Is there any mathematical formula that solves this?
division-ring
edited Jan 4 at 2:11
David G. Stork
9,96021232
asked Jan 4 at 1:51
Eric HondzinskiEric Hondzinski
11
New contributor
Eric Hondzinski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let's say I need to split the following whole number 4 ways: 1659. I cannot have any remainders, yet I need the sum of the four divisors to equal the sum of the dividend. Is there any mathematical formula that solves this?
division-ring
division-ring
edited Jan 4 at 2:11
David G. Stork
9,96021232
asked Jan 4 at 1:51
Eric HondzinskiEric Hondzinski
11
New contributor
Eric Hondzinski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 4 at 2:11
David G. Stork
9,96021232
asked Jan 4 at 1:51
Eric HondzinskiEric Hondzinski
11
New contributor
Eric Hondzinski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 4 at 2:11
David G. Stork
9,96021232
edited Jan 4 at 2:11
David G. Stork
9,96021232
edited Jan 4 at 2:11
David G. Stork
9,96021232
9,96021232
asked Jan 4 at 1:51
Eric HondzinskiEric Hondzinski
11
New contributor
Eric Hondzinski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 1:51
Eric HondzinskiEric Hondzinski
11
asked Jan 4 at 1:51
Eric HondzinskiEric Hondzinski
11
11
New contributor
Eric Hondzinski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Eric Hondzinski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Eric Hondzinski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as unclear what you're asking by Hans Lundmark, user91500, Paul Frost, Cesareo, mrtaurho Jan 4 at 13:34
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by Hans Lundmark, user91500, Paul Frost, Cesareo, mrtaurho Jan 4 at 13:34
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
if we add the digits we get $1+6+5+9 = 21,$ suggesting that 3 is a factor. $1659=3cdot 553$ Now look for factors of $553$ next smallest number that might be a factor is 7.
– Doug M
Jan 4 at 1:56
3
Eric, I have no idea what you are talking about. Is there some number other than 1659 for which you can solve your question completely, and edit that information into the question?
– Will Jagy
Jan 4 at 2:03
add a comment |
if we add the digits we get $1+6+5+9 = 21,$ suggesting that 3 is a factor. $1659=3cdot 553$ Now look for factors of $553$ next smallest number that might be a factor is 7.
– Doug M
Jan 4 at 1:56
3
Eric, I have no idea what you are talking about. Is there some number other than 1659 for which you can solve your question completely, and edit that information into the question?
– Will Jagy
Jan 4 at 2:03
if we add the digits we get $1+6+5+9 = 21,$ suggesting that 3 is a factor. $1659=3cdot 553$ Now look for factors of $553$ next smallest number that might be a factor is 7.
– Doug M
Jan 4 at 1:56
if we add the digits we get $1+6+5+9 = 21,$ suggesting that 3 is a factor. $1659=3cdot 553$ Now look for factors of $553$ next smallest number that might be a factor is 7.
– Doug M
Jan 4 at 1:56
3
3
Eric, I have no idea what you are talking about. Is there some number other than 1659 for which you can solve your question completely, and edit that information into the question?
– Will Jagy
Jan 4 at 2:03
Eric, I have no idea what you are talking about. Is there some number other than 1659 for which you can solve your question completely, and edit that information into the question?
– Will Jagy
Jan 4 at 2:03
add a comment |
1 Answer
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If I understand the question correctly, let us say there is a number $n$ to "split" $m$ ways so that, if the numbers used are $f_i$ for $i = 1, 2, ldots , m$ which are among the factors of $n$, then
$$f_i ; | ; n ; forall ; i = 1, 2, ldots , m ; text{ and } ; sum_{i = 1}^{m} f_i = n tag{1}label{eq1} $$
I don't believe there is any closed mathematical formula to solve this in general. However, you can certainly check for specific cases. In your example, $n = 1659$ and $m = 4$. First, find the prime factors of $1659$ as Doug M suggested in his comment. They are $3$, $7$ and $79$. Thus, the possible values for $f_i$ are all of the combinations of these $3$ values, i.e., $1, 3, 7, 79, 21, 237, 553, 1659$. Since $1659$ can't be used as it would make the sum too large, the terms must all be from the other factors. However, the sum of all of these other factors is $901$, so there is no such situation in your example.
Actually, this brings up a set of values you can exclude. Note that if $m = 1$, the solution is just $n$. However, if $m gt 1$, then the factor of $n$ itself cannot be used, so the sum of the other terms must be $ge n$ for a subset of them to possibly be the solution. Thus deficient numbers (which your value of $1659$ is an example), i.e., where the sum is $lt n$, will not solve it. Perfect numbers, where the sum is $= n$, will only work if $m$ is the exact number of factors less than $n$. More generally, abundant numbers, where the sum is $gt n$, may possibly provide a solution. However, note that relatively few such cases will work. Nonetheless, a simple example using the first abundant number, $12$, is that for $m = 4$, the values are $1, 2, 3, 6$.
answered Jan 4 at 2:14
John OmielanJohn Omielan
1,13918
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If I understand the question correctly, let us say there is a number $n$ to "split" $m$ ways so that, if the numbers used are $f_i$ for $i = 1, 2, ldots , m$ which are among the factors of $n$, then
$$f_i ; | ; n ; forall ; i = 1, 2, ldots , m ; text{ and } ; sum_{i = 1}^{m} f_i = n tag{1}label{eq1} $$
I don't believe there is any closed mathematical formula to solve this in general. However, you can certainly check for specific cases. In your example, $n = 1659$ and $m = 4$. First, find the prime factors of $1659$ as Doug M suggested in his comment. They are $3$, $7$ and $79$. Thus, the possible values for $f_i$ are all of the combinations of these $3$ values, i.e., $1, 3, 7, 79, 21, 237, 553, 1659$. Since $1659$ can't be used as it would make the sum too large, the terms must all be from the other factors. However, the sum of all of these other factors is $901$, so there is no such situation in your example.
Actually, this brings up a set of values you can exclude. Note that if $m = 1$, the solution is just $n$. However, if $m gt 1$, then the factor of $n$ itself cannot be used, so the sum of the other terms must be $ge n$ for a subset of them to possibly be the solution. Thus deficient numbers (which your value of $1659$ is an example), i.e., where the sum is $lt n$, will not solve it. Perfect numbers, where the sum is $= n$, will only work if $m$ is the exact number of factors less than $n$. More generally, abundant numbers, where the sum is $gt n$, may possibly provide a solution. However, note that relatively few such cases will work. Nonetheless, a simple example using the first abundant number, $12$, is that for $m = 4$, the values are $1, 2, 3, 6$.
answered Jan 4 at 2:14
John OmielanJohn Omielan
1,13918
add a comment |
If I understand the question correctly, let us say there is a number $n$ to "split" $m$ ways so that, if the numbers used are $f_i$ for $i = 1, 2, ldots , m$ which are among the factors of $n$, then
$$f_i ; | ; n ; forall ; i = 1, 2, ldots , m ; text{ and } ; sum_{i = 1}^{m} f_i = n tag{1}label{eq1} $$
I don't believe there is any closed mathematical formula to solve this in general. However, you can certainly check for specific cases. In your example, $n = 1659$ and $m = 4$. First, find the prime factors of $1659$ as Doug M suggested in his comment. They are $3$, $7$ and $79$. Thus, the possible values for $f_i$ are all of the combinations of these $3$ values, i.e., $1, 3, 7, 79, 21, 237, 553, 1659$. Since $1659$ can't be used as it would make the sum too large, the terms must all be from the other factors. However, the sum of all of these other factors is $901$, so there is no such situation in your example.
Actually, this brings up a set of values you can exclude. Note that if $m = 1$, the solution is just $n$. However, if $m gt 1$, then the factor of $n$ itself cannot be used, so the sum of the other terms must be $ge n$ for a subset of them to possibly be the solution. Thus deficient numbers (which your value of $1659$ is an example), i.e., where the sum is $lt n$, will not solve it. Perfect numbers, where the sum is $= n$, will only work if $m$ is the exact number of factors less than $n$. More generally, abundant numbers, where the sum is $gt n$, may possibly provide a solution. However, note that relatively few such cases will work. Nonetheless, a simple example using the first abundant number, $12$, is that for $m = 4$, the values are $1, 2, 3, 6$.
answered Jan 4 at 2:14
John OmielanJohn Omielan
1,13918
add a comment |
If I understand the question correctly, let us say there is a number $n$ to "split" $m$ ways so that, if the numbers used are $f_i$ for $i = 1, 2, ldots , m$ which are among the factors of $n$, then
$$f_i ; | ; n ; forall ; i = 1, 2, ldots , m ; text{ and } ; sum_{i = 1}^{m} f_i = n tag{1}label{eq1} $$
I don't believe there is any closed mathematical formula to solve this in general. However, you can certainly check for specific cases. In your example, $n = 1659$ and $m = 4$. First, find the prime factors of $1659$ as Doug M suggested in his comment. They are $3$, $7$ and $79$. Thus, the possible values for $f_i$ are all of the combinations of these $3$ values, i.e., $1, 3, 7, 79, 21, 237, 553, 1659$. Since $1659$ can't be used as it would make the sum too large, the terms must all be from the other factors. However, the sum of all of these other factors is $901$, so there is no such situation in your example.
Actually, this brings up a set of values you can exclude. Note that if $m = 1$, the solution is just $n$. However, if $m gt 1$, then the factor of $n$ itself cannot be used, so the sum of the other terms must be $ge n$ for a subset of them to possibly be the solution. Thus deficient numbers (which your value of $1659$ is an example), i.e., where the sum is $lt n$, will not solve it. Perfect numbers, where the sum is $= n$, will only work if $m$ is the exact number of factors less than $n$. More generally, abundant numbers, where the sum is $gt n$, may possibly provide a solution. However, note that relatively few such cases will work. Nonetheless, a simple example using the first abundant number, $12$, is that for $m = 4$, the values are $1, 2, 3, 6$.
answered Jan 4 at 2:14
John OmielanJohn Omielan
1,13918
If I understand the question correctly, let us say there is a number $n$ to "split" $m$ ways so that, if the numbers used are $f_i$ for $i = 1, 2, ldots , m$ which are among the factors of $n$, then
$$f_i ; | ; n ; forall ; i = 1, 2, ldots , m ; text{ and } ; sum_{i = 1}^{m} f_i = n tag{1}label{eq1} $$
I don't believe there is any closed mathematical formula to solve this in general. However, you can certainly check for specific cases. In your example, $n = 1659$ and $m = 4$. First, find the prime factors of $1659$ as Doug M suggested in his comment. They are $3$, $7$ and $79$. Thus, the possible values for $f_i$ are all of the combinations of these $3$ values, i.e., $1, 3, 7, 79, 21, 237, 553, 1659$. Since $1659$ can't be used as it would make the sum too large, the terms must all be from the other factors. However, the sum of all of these other factors is $901$, so there is no such situation in your example.
Actually, this brings up a set of values you can exclude. Note that if $m = 1$, the solution is just $n$. However, if $m gt 1$, then the factor of $n$ itself cannot be used, so the sum of the other terms must be $ge n$ for a subset of them to possibly be the solution. Thus deficient numbers (which your value of $1659$ is an example), i.e., where the sum is $lt n$, will not solve it. Perfect numbers, where the sum is $= n$, will only work if $m$ is the exact number of factors less than $n$. More generally, abundant numbers, where the sum is $gt n$, may possibly provide a solution. However, note that relatively few such cases will work. Nonetheless, a simple example using the first abundant number, $12$, is that for $m = 4$, the values are $1, 2, 3, 6$.
answered Jan 4 at 2:14
John OmielanJohn Omielan
1,13918
edited Jan 4 at 2:41
answered Jan 4 at 2:14
John OmielanJohn Omielan
1,13918
answered Jan 4 at 2:14
John OmielanJohn Omielan
1,13918
answered Jan 4 at 2:14
John OmielanJohn Omielan
1,13918
1,13918
add a comment |
add a comment |
if we add the digits we get $1+6+5+9 = 21,$ suggesting that 3 is a factor. $1659=3cdot 553$ Now look for factors of $553$ next smallest number that might be a factor is 7.
– Doug M
Jan 4 at 1:56
3
Eric, I have no idea what you are talking about. Is there some number other than 1659 for which you can solve your question completely, and edit that information into the question?
– Will Jagy
Jan 4 at 2:03