I need an expression that equals 113079 [closed]

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1
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My brother passed away last year. I want to get a tattoo of his initials JWJ equaling an expression that equals his birthday 113079. He was a math whiz and thought this would be a cool idea for a tattoo for him.
algebra-precalculus
closed as off-topic by Peter, Xander Henderson, Jendrik Stelzner, Tyrone, GNU Supporter Aug 10 at 9:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is not about mathematics, within the scope defined in the help center." â Xander Henderson, Jendrik Stelzner, Tyrone
add a comment |Â
up vote
1
down vote
favorite
My brother passed away last year. I want to get a tattoo of his initials JWJ equaling an expression that equals his birthday 113079. He was a math whiz and thought this would be a cool idea for a tattoo for him.
algebra-precalculus
closed as off-topic by Peter, Xander Henderson, Jendrik Stelzner, Tyrone, GNU Supporter Aug 10 at 9:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is not about mathematics, within the scope defined in the help center." â Xander Henderson, Jendrik Stelzner, Tyrone
1
113079 = 113079. Perhaps you'd like to specify what kind of expression you want?
â John Dvorak
Aug 9 at 15:39
5
Did he have a favorite kind of math? Geometry, calculus, etc.?
â Clayton
Aug 9 at 15:40
Prime factorization: $3 times 37693$.
â David G. Stork
Aug 9 at 15:59
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My brother passed away last year. I want to get a tattoo of his initials JWJ equaling an expression that equals his birthday 113079. He was a math whiz and thought this would be a cool idea for a tattoo for him.
algebra-precalculus
My brother passed away last year. I want to get a tattoo of his initials JWJ equaling an expression that equals his birthday 113079. He was a math whiz and thought this would be a cool idea for a tattoo for him.
algebra-precalculus
edited Aug 9 at 16:23
amWhy
189k25219431
189k25219431
asked Aug 9 at 15:37
JenaJ
171
171
closed as off-topic by Peter, Xander Henderson, Jendrik Stelzner, Tyrone, GNU Supporter Aug 10 at 9:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is not about mathematics, within the scope defined in the help center." â Xander Henderson, Jendrik Stelzner, Tyrone
closed as off-topic by Peter, Xander Henderson, Jendrik Stelzner, Tyrone, GNU Supporter Aug 10 at 9:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is not about mathematics, within the scope defined in the help center." â Xander Henderson, Jendrik Stelzner, Tyrone
1
113079 = 113079. Perhaps you'd like to specify what kind of expression you want?
â John Dvorak
Aug 9 at 15:39
5
Did he have a favorite kind of math? Geometry, calculus, etc.?
â Clayton
Aug 9 at 15:40
Prime factorization: $3 times 37693$.
â David G. Stork
Aug 9 at 15:59
add a comment |Â
1
113079 = 113079. Perhaps you'd like to specify what kind of expression you want?
â John Dvorak
Aug 9 at 15:39
5
Did he have a favorite kind of math? Geometry, calculus, etc.?
â Clayton
Aug 9 at 15:40
Prime factorization: $3 times 37693$.
â David G. Stork
Aug 9 at 15:59
1
1
113079 = 113079. Perhaps you'd like to specify what kind of expression you want?
â John Dvorak
Aug 9 at 15:39
113079 = 113079. Perhaps you'd like to specify what kind of expression you want?
â John Dvorak
Aug 9 at 15:39
5
5
Did he have a favorite kind of math? Geometry, calculus, etc.?
â Clayton
Aug 9 at 15:40
Did he have a favorite kind of math? Geometry, calculus, etc.?
â Clayton
Aug 9 at 15:40
Prime factorization: $3 times 37693$.
â David G. Stork
Aug 9 at 15:59
Prime factorization: $3 times 37693$.
â David G. Stork
Aug 9 at 15:59
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
Here are some ideas that make vague mathematical sense!
$$sum_j=0^infty w_j=113079$$
$$j^3+w^3+J^3=113079$$
$$J(w(j))=113079$$
$$J^11+W^30=J^79$$
$$mathbfjwedge,vec WJ=left( beginmatrix 11 \ 30 \ 79endmatrix right)$$
add a comment |Â
up vote
1
down vote
There are lots of expressions which equal that, unfortunately. I guess you're looking for some sort of interesting one, for which I'd suggest the prime factorization.
Unfortunately, the prime factorisation is rather boring:
$3 times 37693$
EDIT:
However, you could do
$J.W.J. = 2 times 7 times 41 times 197 + 1 $
or
$ J.W.J. = 2 times 2 times 2 times 5 times 11 times 257 -1 $
Finally, there's also stuff more like this:
$$fracprod_n=1^11n300 - sum_n=1^796n - 1017$$
which looks a bit more "equationy".
I really like this one that you call "equationy". Thank you.
â JenaJ
Aug 10 at 18:35
You're welcome. If you do decide to use this, and want a heads up on exactly how it should be written/any other details, please feel free to comment here and I'll elaborate.
â Matt
Aug 10 at 21:00
It's definitely the one I would like to use. It's more than mere coincidence that he had the pi symbol tattooed on his wrist and my wrist is the location I chose.
â JenaJ
Aug 12 at 22:00
When are you planning to have it done? I think the best bet would be to get a print out of the above, and get the artist to simply trace it. With a formula like that, the most minor slip can make the whole thing wrong, or not even make sense, and I'm sure you want it to be right :-)
â Matt
Aug 12 at 22:23
I'm planning on going this weekend or next week. This Thursday will be one year since he passed. And good idea on getting it printed out, it needs to be perfect. Thank you for the help, I know others were not very pleased with my post on this site but when you love someone you do whatever it takes.
â JenaJ
Aug 13 at 23:51
add a comment |Â
up vote
1
down vote
This number is the sum of three cubes, i.e.,
$$
113079=J^3+W^3+J'^3,
$$
with (large) integers $J,W,J'$.
Reference: This MO-question.
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
â Brahadeesh
Aug 9 at 20:10
@Brahadeesh But Malkin independently gave the same answer above. Why don't you comment on his answer then? And indeed, $113079$ is the sum of $3$ cubes, answering the title question.
â Dietrich Burde
Aug 10 at 8:51
Actually, it is easier with $4$ cubes, e.g., $113079=(-37688)^3+37687^3+18846^3+(-18842)^3$, but we only have $3$ initials for the name, of course.
â Dietrich Burde
Aug 10 at 9:10
Yikes, I misunderstood your answer completely. My apologies, IâÂÂll be more careful from now on.
â Brahadeesh
Aug 10 at 10:07
Thank you all so much. I'm sorry if this post wasn't what this site is for but I thought I would try. Trigonometry was his favorite.
â JenaJ
Aug 10 at 18:33
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Here are some ideas that make vague mathematical sense!
$$sum_j=0^infty w_j=113079$$
$$j^3+w^3+J^3=113079$$
$$J(w(j))=113079$$
$$J^11+W^30=J^79$$
$$mathbfjwedge,vec WJ=left( beginmatrix 11 \ 30 \ 79endmatrix right)$$
add a comment |Â
up vote
2
down vote
Here are some ideas that make vague mathematical sense!
$$sum_j=0^infty w_j=113079$$
$$j^3+w^3+J^3=113079$$
$$J(w(j))=113079$$
$$J^11+W^30=J^79$$
$$mathbfjwedge,vec WJ=left( beginmatrix 11 \ 30 \ 79endmatrix right)$$
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Here are some ideas that make vague mathematical sense!
$$sum_j=0^infty w_j=113079$$
$$j^3+w^3+J^3=113079$$
$$J(w(j))=113079$$
$$J^11+W^30=J^79$$
$$mathbfjwedge,vec WJ=left( beginmatrix 11 \ 30 \ 79endmatrix right)$$
Here are some ideas that make vague mathematical sense!
$$sum_j=0^infty w_j=113079$$
$$j^3+w^3+J^3=113079$$
$$J(w(j))=113079$$
$$J^11+W^30=J^79$$
$$mathbfjwedge,vec WJ=left( beginmatrix 11 \ 30 \ 79endmatrix right)$$
answered Aug 9 at 16:11
Malkin
1,482523
1,482523
add a comment |Â
add a comment |Â
up vote
1
down vote
There are lots of expressions which equal that, unfortunately. I guess you're looking for some sort of interesting one, for which I'd suggest the prime factorization.
Unfortunately, the prime factorisation is rather boring:
$3 times 37693$
EDIT:
However, you could do
$J.W.J. = 2 times 7 times 41 times 197 + 1 $
or
$ J.W.J. = 2 times 2 times 2 times 5 times 11 times 257 -1 $
Finally, there's also stuff more like this:
$$fracprod_n=1^11n300 - sum_n=1^796n - 1017$$
which looks a bit more "equationy".
I really like this one that you call "equationy". Thank you.
â JenaJ
Aug 10 at 18:35
You're welcome. If you do decide to use this, and want a heads up on exactly how it should be written/any other details, please feel free to comment here and I'll elaborate.
â Matt
Aug 10 at 21:00
It's definitely the one I would like to use. It's more than mere coincidence that he had the pi symbol tattooed on his wrist and my wrist is the location I chose.
â JenaJ
Aug 12 at 22:00
When are you planning to have it done? I think the best bet would be to get a print out of the above, and get the artist to simply trace it. With a formula like that, the most minor slip can make the whole thing wrong, or not even make sense, and I'm sure you want it to be right :-)
â Matt
Aug 12 at 22:23
I'm planning on going this weekend or next week. This Thursday will be one year since he passed. And good idea on getting it printed out, it needs to be perfect. Thank you for the help, I know others were not very pleased with my post on this site but when you love someone you do whatever it takes.
â JenaJ
Aug 13 at 23:51
add a comment |Â
up vote
1
down vote
There are lots of expressions which equal that, unfortunately. I guess you're looking for some sort of interesting one, for which I'd suggest the prime factorization.
Unfortunately, the prime factorisation is rather boring:
$3 times 37693$
EDIT:
However, you could do
$J.W.J. = 2 times 7 times 41 times 197 + 1 $
or
$ J.W.J. = 2 times 2 times 2 times 5 times 11 times 257 -1 $
Finally, there's also stuff more like this:
$$fracprod_n=1^11n300 - sum_n=1^796n - 1017$$
which looks a bit more "equationy".
I really like this one that you call "equationy". Thank you.
â JenaJ
Aug 10 at 18:35
You're welcome. If you do decide to use this, and want a heads up on exactly how it should be written/any other details, please feel free to comment here and I'll elaborate.
â Matt
Aug 10 at 21:00
It's definitely the one I would like to use. It's more than mere coincidence that he had the pi symbol tattooed on his wrist and my wrist is the location I chose.
â JenaJ
Aug 12 at 22:00
When are you planning to have it done? I think the best bet would be to get a print out of the above, and get the artist to simply trace it. With a formula like that, the most minor slip can make the whole thing wrong, or not even make sense, and I'm sure you want it to be right :-)
â Matt
Aug 12 at 22:23
I'm planning on going this weekend or next week. This Thursday will be one year since he passed. And good idea on getting it printed out, it needs to be perfect. Thank you for the help, I know others were not very pleased with my post on this site but when you love someone you do whatever it takes.
â JenaJ
Aug 13 at 23:51
add a comment |Â
up vote
1
down vote
up vote
1
down vote
There are lots of expressions which equal that, unfortunately. I guess you're looking for some sort of interesting one, for which I'd suggest the prime factorization.
Unfortunately, the prime factorisation is rather boring:
$3 times 37693$
EDIT:
However, you could do
$J.W.J. = 2 times 7 times 41 times 197 + 1 $
or
$ J.W.J. = 2 times 2 times 2 times 5 times 11 times 257 -1 $
Finally, there's also stuff more like this:
$$fracprod_n=1^11n300 - sum_n=1^796n - 1017$$
which looks a bit more "equationy".
There are lots of expressions which equal that, unfortunately. I guess you're looking for some sort of interesting one, for which I'd suggest the prime factorization.
Unfortunately, the prime factorisation is rather boring:
$3 times 37693$
EDIT:
However, you could do
$J.W.J. = 2 times 7 times 41 times 197 + 1 $
or
$ J.W.J. = 2 times 2 times 2 times 5 times 11 times 257 -1 $
Finally, there's also stuff more like this:
$$fracprod_n=1^11n300 - sum_n=1^796n - 1017$$
which looks a bit more "equationy".
edited Aug 9 at 16:17
answered Aug 9 at 15:42
Matt
1,972617
1,972617
I really like this one that you call "equationy". Thank you.
â JenaJ
Aug 10 at 18:35
You're welcome. If you do decide to use this, and want a heads up on exactly how it should be written/any other details, please feel free to comment here and I'll elaborate.
â Matt
Aug 10 at 21:00
It's definitely the one I would like to use. It's more than mere coincidence that he had the pi symbol tattooed on his wrist and my wrist is the location I chose.
â JenaJ
Aug 12 at 22:00
When are you planning to have it done? I think the best bet would be to get a print out of the above, and get the artist to simply trace it. With a formula like that, the most minor slip can make the whole thing wrong, or not even make sense, and I'm sure you want it to be right :-)
â Matt
Aug 12 at 22:23
I'm planning on going this weekend or next week. This Thursday will be one year since he passed. And good idea on getting it printed out, it needs to be perfect. Thank you for the help, I know others were not very pleased with my post on this site but when you love someone you do whatever it takes.
â JenaJ
Aug 13 at 23:51
add a comment |Â
I really like this one that you call "equationy". Thank you.
â JenaJ
Aug 10 at 18:35
You're welcome. If you do decide to use this, and want a heads up on exactly how it should be written/any other details, please feel free to comment here and I'll elaborate.
â Matt
Aug 10 at 21:00
It's definitely the one I would like to use. It's more than mere coincidence that he had the pi symbol tattooed on his wrist and my wrist is the location I chose.
â JenaJ
Aug 12 at 22:00
When are you planning to have it done? I think the best bet would be to get a print out of the above, and get the artist to simply trace it. With a formula like that, the most minor slip can make the whole thing wrong, or not even make sense, and I'm sure you want it to be right :-)
â Matt
Aug 12 at 22:23
I'm planning on going this weekend or next week. This Thursday will be one year since he passed. And good idea on getting it printed out, it needs to be perfect. Thank you for the help, I know others were not very pleased with my post on this site but when you love someone you do whatever it takes.
â JenaJ
Aug 13 at 23:51
I really like this one that you call "equationy". Thank you.
â JenaJ
Aug 10 at 18:35
I really like this one that you call "equationy". Thank you.
â JenaJ
Aug 10 at 18:35
You're welcome. If you do decide to use this, and want a heads up on exactly how it should be written/any other details, please feel free to comment here and I'll elaborate.
â Matt
Aug 10 at 21:00
You're welcome. If you do decide to use this, and want a heads up on exactly how it should be written/any other details, please feel free to comment here and I'll elaborate.
â Matt
Aug 10 at 21:00
It's definitely the one I would like to use. It's more than mere coincidence that he had the pi symbol tattooed on his wrist and my wrist is the location I chose.
â JenaJ
Aug 12 at 22:00
It's definitely the one I would like to use. It's more than mere coincidence that he had the pi symbol tattooed on his wrist and my wrist is the location I chose.
â JenaJ
Aug 12 at 22:00
When are you planning to have it done? I think the best bet would be to get a print out of the above, and get the artist to simply trace it. With a formula like that, the most minor slip can make the whole thing wrong, or not even make sense, and I'm sure you want it to be right :-)
â Matt
Aug 12 at 22:23
When are you planning to have it done? I think the best bet would be to get a print out of the above, and get the artist to simply trace it. With a formula like that, the most minor slip can make the whole thing wrong, or not even make sense, and I'm sure you want it to be right :-)
â Matt
Aug 12 at 22:23
I'm planning on going this weekend or next week. This Thursday will be one year since he passed. And good idea on getting it printed out, it needs to be perfect. Thank you for the help, I know others were not very pleased with my post on this site but when you love someone you do whatever it takes.
â JenaJ
Aug 13 at 23:51
I'm planning on going this weekend or next week. This Thursday will be one year since he passed. And good idea on getting it printed out, it needs to be perfect. Thank you for the help, I know others were not very pleased with my post on this site but when you love someone you do whatever it takes.
â JenaJ
Aug 13 at 23:51
add a comment |Â
up vote
1
down vote
This number is the sum of three cubes, i.e.,
$$
113079=J^3+W^3+J'^3,
$$
with (large) integers $J,W,J'$.
Reference: This MO-question.
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
â Brahadeesh
Aug 9 at 20:10
@Brahadeesh But Malkin independently gave the same answer above. Why don't you comment on his answer then? And indeed, $113079$ is the sum of $3$ cubes, answering the title question.
â Dietrich Burde
Aug 10 at 8:51
Actually, it is easier with $4$ cubes, e.g., $113079=(-37688)^3+37687^3+18846^3+(-18842)^3$, but we only have $3$ initials for the name, of course.
â Dietrich Burde
Aug 10 at 9:10
Yikes, I misunderstood your answer completely. My apologies, IâÂÂll be more careful from now on.
â Brahadeesh
Aug 10 at 10:07
Thank you all so much. I'm sorry if this post wasn't what this site is for but I thought I would try. Trigonometry was his favorite.
â JenaJ
Aug 10 at 18:33
add a comment |Â
up vote
1
down vote
This number is the sum of three cubes, i.e.,
$$
113079=J^3+W^3+J'^3,
$$
with (large) integers $J,W,J'$.
Reference: This MO-question.
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
â Brahadeesh
Aug 9 at 20:10
@Brahadeesh But Malkin independently gave the same answer above. Why don't you comment on his answer then? And indeed, $113079$ is the sum of $3$ cubes, answering the title question.
â Dietrich Burde
Aug 10 at 8:51
Actually, it is easier with $4$ cubes, e.g., $113079=(-37688)^3+37687^3+18846^3+(-18842)^3$, but we only have $3$ initials for the name, of course.
â Dietrich Burde
Aug 10 at 9:10
Yikes, I misunderstood your answer completely. My apologies, IâÂÂll be more careful from now on.
â Brahadeesh
Aug 10 at 10:07
Thank you all so much. I'm sorry if this post wasn't what this site is for but I thought I would try. Trigonometry was his favorite.
â JenaJ
Aug 10 at 18:33
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This number is the sum of three cubes, i.e.,
$$
113079=J^3+W^3+J'^3,
$$
with (large) integers $J,W,J'$.
Reference: This MO-question.
This number is the sum of three cubes, i.e.,
$$
113079=J^3+W^3+J'^3,
$$
with (large) integers $J,W,J'$.
Reference: This MO-question.
edited Aug 10 at 8:55
answered Aug 9 at 15:45
Dietrich Burde
74.8k64185
74.8k64185
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
â Brahadeesh
Aug 9 at 20:10
@Brahadeesh But Malkin independently gave the same answer above. Why don't you comment on his answer then? And indeed, $113079$ is the sum of $3$ cubes, answering the title question.
â Dietrich Burde
Aug 10 at 8:51
Actually, it is easier with $4$ cubes, e.g., $113079=(-37688)^3+37687^3+18846^3+(-18842)^3$, but we only have $3$ initials for the name, of course.
â Dietrich Burde
Aug 10 at 9:10
Yikes, I misunderstood your answer completely. My apologies, IâÂÂll be more careful from now on.
â Brahadeesh
Aug 10 at 10:07
Thank you all so much. I'm sorry if this post wasn't what this site is for but I thought I would try. Trigonometry was his favorite.
â JenaJ
Aug 10 at 18:33
add a comment |Â
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
â Brahadeesh
Aug 9 at 20:10
@Brahadeesh But Malkin independently gave the same answer above. Why don't you comment on his answer then? And indeed, $113079$ is the sum of $3$ cubes, answering the title question.
â Dietrich Burde
Aug 10 at 8:51
Actually, it is easier with $4$ cubes, e.g., $113079=(-37688)^3+37687^3+18846^3+(-18842)^3$, but we only have $3$ initials for the name, of course.
â Dietrich Burde
Aug 10 at 9:10
Yikes, I misunderstood your answer completely. My apologies, IâÂÂll be more careful from now on.
â Brahadeesh
Aug 10 at 10:07
Thank you all so much. I'm sorry if this post wasn't what this site is for but I thought I would try. Trigonometry was his favorite.
â JenaJ
Aug 10 at 18:33
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
â Brahadeesh
Aug 9 at 20:10
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
â Brahadeesh
Aug 9 at 20:10
@Brahadeesh But Malkin independently gave the same answer above. Why don't you comment on his answer then? And indeed, $113079$ is the sum of $3$ cubes, answering the title question.
â Dietrich Burde
Aug 10 at 8:51
@Brahadeesh But Malkin independently gave the same answer above. Why don't you comment on his answer then? And indeed, $113079$ is the sum of $3$ cubes, answering the title question.
â Dietrich Burde
Aug 10 at 8:51
Actually, it is easier with $4$ cubes, e.g., $113079=(-37688)^3+37687^3+18846^3+(-18842)^3$, but we only have $3$ initials for the name, of course.
â Dietrich Burde
Aug 10 at 9:10
Actually, it is easier with $4$ cubes, e.g., $113079=(-37688)^3+37687^3+18846^3+(-18842)^3$, but we only have $3$ initials for the name, of course.
â Dietrich Burde
Aug 10 at 9:10
Yikes, I misunderstood your answer completely. My apologies, IâÂÂll be more careful from now on.
â Brahadeesh
Aug 10 at 10:07
Yikes, I misunderstood your answer completely. My apologies, IâÂÂll be more careful from now on.
â Brahadeesh
Aug 10 at 10:07
Thank you all so much. I'm sorry if this post wasn't what this site is for but I thought I would try. Trigonometry was his favorite.
â JenaJ
Aug 10 at 18:33
Thank you all so much. I'm sorry if this post wasn't what this site is for but I thought I would try. Trigonometry was his favorite.
â JenaJ
Aug 10 at 18:33
add a comment |Â
1
113079 = 113079. Perhaps you'd like to specify what kind of expression you want?
â John Dvorak
Aug 9 at 15:39
5
Did he have a favorite kind of math? Geometry, calculus, etc.?
â Clayton
Aug 9 at 15:40
Prime factorization: $3 times 37693$.
â David G. Stork
Aug 9 at 15:59