A crazy number theoretic problem
$begingroup$
How many positive integers satisfy the condition that the product of their digits multiplied by the sum of their digits equals themselves? For example,
13 is not equal to (1+3)(1×3)
Here is how i proceeded
I defined a number n such that $10^{k+1}$$>$$n$$>$$10^k$
Next, i will expand the number in powers of tens.
Now i am trying to create a bound, but at this stage, i am unsuccessful in each of my attempts. Please help me!
number-theory
$endgroup$
|
show 7 more comments
$begingroup$
How many positive integers satisfy the condition that the product of their digits multiplied by the sum of their digits equals themselves? For example,
13 is not equal to (1+3)(1×3)
Here is how i proceeded
I defined a number n such that $10^{k+1}$$>$$n$$>$$10^k$
Next, i will expand the number in powers of tens.
Now i am trying to create a bound, but at this stage, i am unsuccessful in each of my attempts. Please help me!
number-theory
$endgroup$
1
$begingroup$
Did you search the site this time before posting?
$endgroup$
– Dietrich Burde
Jan 21 at 17:26
$begingroup$
What do you mean?
$endgroup$
– user636268
Jan 21 at 17:26
$begingroup$
I mean, whether your question is new here or already has answers, i.e., is a duplicate?
$endgroup$
– Dietrich Burde
Jan 21 at 17:27
6
$begingroup$
There's actually a name for this sort of number! Let $n$ be a number such that $$n = text{(product of n's digits)} times text{(sum of n's digits)}$$ Then $n$ is what we call a "sum-product number." (en.wikipedia.org/wiki/Sum-product_number) In base $10$, there are only four such numbers, only two of which are nontrivial - 135 and 144. It has been proven that the number of such numbers in any given base is also finite. The Wikipedia article linked will be a good starting point for looking into this topic. I don't know much offhand about proving stuff of this sort though.
$endgroup$
– Eevee Trainer
Jan 21 at 17:55
2
$begingroup$
"But i assume this question will not have any such duplicates" No. As a rule of thumb, every idea you have is unoriginal.
$endgroup$
– Servaes
Jan 21 at 18:07
|
show 7 more comments
$begingroup$
How many positive integers satisfy the condition that the product of their digits multiplied by the sum of their digits equals themselves? For example,
13 is not equal to (1+3)(1×3)
Here is how i proceeded
I defined a number n such that $10^{k+1}$$>$$n$$>$$10^k$
Next, i will expand the number in powers of tens.
Now i am trying to create a bound, but at this stage, i am unsuccessful in each of my attempts. Please help me!
number-theory
$endgroup$
How many positive integers satisfy the condition that the product of their digits multiplied by the sum of their digits equals themselves? For example,
13 is not equal to (1+3)(1×3)
Here is how i proceeded
I defined a number n such that $10^{k+1}$$>$$n$$>$$10^k$
Next, i will expand the number in powers of tens.
Now i am trying to create a bound, but at this stage, i am unsuccessful in each of my attempts. Please help me!
number-theory
number-theory
asked Jan 21 at 17:24
user636268
1
$begingroup$
Did you search the site this time before posting?
$endgroup$
– Dietrich Burde
Jan 21 at 17:26
$begingroup$
What do you mean?
$endgroup$
– user636268
Jan 21 at 17:26
$begingroup$
I mean, whether your question is new here or already has answers, i.e., is a duplicate?
$endgroup$
– Dietrich Burde
Jan 21 at 17:27
6
$begingroup$
There's actually a name for this sort of number! Let $n$ be a number such that $$n = text{(product of n's digits)} times text{(sum of n's digits)}$$ Then $n$ is what we call a "sum-product number." (en.wikipedia.org/wiki/Sum-product_number) In base $10$, there are only four such numbers, only two of which are nontrivial - 135 and 144. It has been proven that the number of such numbers in any given base is also finite. The Wikipedia article linked will be a good starting point for looking into this topic. I don't know much offhand about proving stuff of this sort though.
$endgroup$
– Eevee Trainer
Jan 21 at 17:55
2
$begingroup$
"But i assume this question will not have any such duplicates" No. As a rule of thumb, every idea you have is unoriginal.
$endgroup$
– Servaes
Jan 21 at 18:07
|
show 7 more comments
1
$begingroup$
Did you search the site this time before posting?
$endgroup$
– Dietrich Burde
Jan 21 at 17:26
$begingroup$
What do you mean?
$endgroup$
– user636268
Jan 21 at 17:26
$begingroup$
I mean, whether your question is new here or already has answers, i.e., is a duplicate?
$endgroup$
– Dietrich Burde
Jan 21 at 17:27
6
$begingroup$
There's actually a name for this sort of number! Let $n$ be a number such that $$n = text{(product of n's digits)} times text{(sum of n's digits)}$$ Then $n$ is what we call a "sum-product number." (en.wikipedia.org/wiki/Sum-product_number) In base $10$, there are only four such numbers, only two of which are nontrivial - 135 and 144. It has been proven that the number of such numbers in any given base is also finite. The Wikipedia article linked will be a good starting point for looking into this topic. I don't know much offhand about proving stuff of this sort though.
$endgroup$
– Eevee Trainer
Jan 21 at 17:55
2
$begingroup$
"But i assume this question will not have any such duplicates" No. As a rule of thumb, every idea you have is unoriginal.
$endgroup$
– Servaes
Jan 21 at 18:07
1
1
$begingroup$
Did you search the site this time before posting?
$endgroup$
– Dietrich Burde
Jan 21 at 17:26
$begingroup$
Did you search the site this time before posting?
$endgroup$
– Dietrich Burde
Jan 21 at 17:26
$begingroup$
What do you mean?
$endgroup$
– user636268
Jan 21 at 17:26
$begingroup$
What do you mean?
$endgroup$
– user636268
Jan 21 at 17:26
$begingroup$
I mean, whether your question is new here or already has answers, i.e., is a duplicate?
$endgroup$
– Dietrich Burde
Jan 21 at 17:27
$begingroup$
I mean, whether your question is new here or already has answers, i.e., is a duplicate?
$endgroup$
– Dietrich Burde
Jan 21 at 17:27
6
6
$begingroup$
There's actually a name for this sort of number! Let $n$ be a number such that $$n = text{(product of n's digits)} times text{(sum of n's digits)}$$ Then $n$ is what we call a "sum-product number." (en.wikipedia.org/wiki/Sum-product_number) In base $10$, there are only four such numbers, only two of which are nontrivial - 135 and 144. It has been proven that the number of such numbers in any given base is also finite. The Wikipedia article linked will be a good starting point for looking into this topic. I don't know much offhand about proving stuff of this sort though.
$endgroup$
– Eevee Trainer
Jan 21 at 17:55
$begingroup$
There's actually a name for this sort of number! Let $n$ be a number such that $$n = text{(product of n's digits)} times text{(sum of n's digits)}$$ Then $n$ is what we call a "sum-product number." (en.wikipedia.org/wiki/Sum-product_number) In base $10$, there are only four such numbers, only two of which are nontrivial - 135 and 144. It has been proven that the number of such numbers in any given base is also finite. The Wikipedia article linked will be a good starting point for looking into this topic. I don't know much offhand about proving stuff of this sort though.
$endgroup$
– Eevee Trainer
Jan 21 at 17:55
2
2
$begingroup$
"But i assume this question will not have any such duplicates" No. As a rule of thumb, every idea you have is unoriginal.
$endgroup$
– Servaes
Jan 21 at 18:07
$begingroup$
"But i assume this question will not have any such duplicates" No. As a rule of thumb, every idea you have is unoriginal.
$endgroup$
– Servaes
Jan 21 at 18:07
|
show 7 more comments
1 Answer
1
active
oldest
votes
$begingroup$
To set an upper bound, note that each digit is at most $9$. If the number has $n$ digits, the computation gives at most $(9n)(9^n)=n9^{n+1}$. An $n$ digit number is at least $10^{n-1}$. Alpha tells us that for $n=85$ we have the number is greater than the sum of digits times the product of digits, so you only have to search up to $10^{84}$. Happy hunting.
Now note that most of the factors of the number need to be $2,3,5,7$ because the only way you can get a larger one is from the sum of the digits. As lulu points out you can't have both $2$ and $5$ as factors because it would end in zero and the product would be zero. The number needs to be $2^a3^b7^cd$ or $3^a5^b7^cd$ with $d lt 9cdot 84$. If you have a programming language that supports large ints, like Python, and a nice way to turn a number into a string you can just do the search. A little cleverness can speed it up, like noting that $e$ can't be that large unless the other numbers are large.
answered Jan 21 at 17:35
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
So whats next? You just created the bound, but this is just a mere observation
$endgroup$
– user636268
Jan 21 at 17:49
1
$begingroup$
@Md.ShamimAkhtar You could give it some thought yourself. You might learn something from it.
$endgroup$
– Servaes
Jan 21 at 18:09
1
$begingroup$
A bound is what you asked for.
$endgroup$
– Ross Millikan
Jan 21 at 18:15
1
$begingroup$
Note: saves time to note that your number can't be divisible by both $2$ and $5$ (else it would end in $0$ and the product of its digits would be $0$).
$endgroup$
– lulu
Jan 21 at 19:04
1
$begingroup$
I don't think this is a reasonable contest problem unless you can reduce the search space much more. I don't know how and the Wikipedia article only cites the same upper limit I found.
$endgroup$
– Ross Millikan
Jan 22 at 2:17
|
show 2 more comments
Your Answer
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1 Answer
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oldest
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1 Answer
1
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oldest
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active
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active
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votes
$begingroup$
To set an upper bound, note that each digit is at most $9$. If the number has $n$ digits, the computation gives at most $(9n)(9^n)=n9^{n+1}$. An $n$ digit number is at least $10^{n-1}$. Alpha tells us that for $n=85$ we have the number is greater than the sum of digits times the product of digits, so you only have to search up to $10^{84}$. Happy hunting.
Now note that most of the factors of the number need to be $2,3,5,7$ because the only way you can get a larger one is from the sum of the digits. As lulu points out you can't have both $2$ and $5$ as factors because it would end in zero and the product would be zero. The number needs to be $2^a3^b7^cd$ or $3^a5^b7^cd$ with $d lt 9cdot 84$. If you have a programming language that supports large ints, like Python, and a nice way to turn a number into a string you can just do the search. A little cleverness can speed it up, like noting that $e$ can't be that large unless the other numbers are large.
answered Jan 21 at 17:35
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
So whats next? You just created the bound, but this is just a mere observation
$endgroup$
– user636268
Jan 21 at 17:49
1
$begingroup$
@Md.ShamimAkhtar You could give it some thought yourself. You might learn something from it.
$endgroup$
– Servaes
Jan 21 at 18:09
1
$begingroup$
A bound is what you asked for.
$endgroup$
– Ross Millikan
Jan 21 at 18:15
1
$begingroup$
Note: saves time to note that your number can't be divisible by both $2$ and $5$ (else it would end in $0$ and the product of its digits would be $0$).
$endgroup$
– lulu
Jan 21 at 19:04
1
$begingroup$
I don't think this is a reasonable contest problem unless you can reduce the search space much more. I don't know how and the Wikipedia article only cites the same upper limit I found.
$endgroup$
– Ross Millikan
Jan 22 at 2:17
|
show 2 more comments
$begingroup$
To set an upper bound, note that each digit is at most $9$. If the number has $n$ digits, the computation gives at most $(9n)(9^n)=n9^{n+1}$. An $n$ digit number is at least $10^{n-1}$. Alpha tells us that for $n=85$ we have the number is greater than the sum of digits times the product of digits, so you only have to search up to $10^{84}$. Happy hunting.
Now note that most of the factors of the number need to be $2,3,5,7$ because the only way you can get a larger one is from the sum of the digits. As lulu points out you can't have both $2$ and $5$ as factors because it would end in zero and the product would be zero. The number needs to be $2^a3^b7^cd$ or $3^a5^b7^cd$ with $d lt 9cdot 84$. If you have a programming language that supports large ints, like Python, and a nice way to turn a number into a string you can just do the search. A little cleverness can speed it up, like noting that $e$ can't be that large unless the other numbers are large.
answered Jan 21 at 17:35
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
So whats next? You just created the bound, but this is just a mere observation
$endgroup$
– user636268
Jan 21 at 17:49
1
$begingroup$
@Md.ShamimAkhtar You could give it some thought yourself. You might learn something from it.
$endgroup$
– Servaes
Jan 21 at 18:09
1
$begingroup$
A bound is what you asked for.
$endgroup$
– Ross Millikan
Jan 21 at 18:15
1
$begingroup$
Note: saves time to note that your number can't be divisible by both $2$ and $5$ (else it would end in $0$ and the product of its digits would be $0$).
$endgroup$
– lulu
Jan 21 at 19:04
1
$begingroup$
I don't think this is a reasonable contest problem unless you can reduce the search space much more. I don't know how and the Wikipedia article only cites the same upper limit I found.
$endgroup$
– Ross Millikan
Jan 22 at 2:17
|
show 2 more comments
$begingroup$
To set an upper bound, note that each digit is at most $9$. If the number has $n$ digits, the computation gives at most $(9n)(9^n)=n9^{n+1}$. An $n$ digit number is at least $10^{n-1}$. Alpha tells us that for $n=85$ we have the number is greater than the sum of digits times the product of digits, so you only have to search up to $10^{84}$. Happy hunting.
Now note that most of the factors of the number need to be $2,3,5,7$ because the only way you can get a larger one is from the sum of the digits. As lulu points out you can't have both $2$ and $5$ as factors because it would end in zero and the product would be zero. The number needs to be $2^a3^b7^cd$ or $3^a5^b7^cd$ with $d lt 9cdot 84$. If you have a programming language that supports large ints, like Python, and a nice way to turn a number into a string you can just do the search. A little cleverness can speed it up, like noting that $e$ can't be that large unless the other numbers are large.
answered Jan 21 at 17:35
Ross MillikanRoss Millikan
297k23198371
$endgroup$
To set an upper bound, note that each digit is at most $9$. If the number has $n$ digits, the computation gives at most $(9n)(9^n)=n9^{n+1}$. An $n$ digit number is at least $10^{n-1}$. Alpha tells us that for $n=85$ we have the number is greater than the sum of digits times the product of digits, so you only have to search up to $10^{84}$. Happy hunting.
Now note that most of the factors of the number need to be $2,3,5,7$ because the only way you can get a larger one is from the sum of the digits. As lulu points out you can't have both $2$ and $5$ as factors because it would end in zero and the product would be zero. The number needs to be $2^a3^b7^cd$ or $3^a5^b7^cd$ with $d lt 9cdot 84$. If you have a programming language that supports large ints, like Python, and a nice way to turn a number into a string you can just do the search. A little cleverness can speed it up, like noting that $e$ can't be that large unless the other numbers are large.
answered Jan 21 at 17:35
Ross MillikanRoss Millikan
297k23198371
edited Jan 21 at 19:09
answered Jan 21 at 17:35
Ross MillikanRoss Millikan
297k23198371
answered Jan 21 at 17:35
Ross MillikanRoss Millikan
297k23198371
answered Jan 21 at 17:35
Ross MillikanRoss Millikan
297k23198371
297k23198371
$begingroup$
So whats next? You just created the bound, but this is just a mere observation
$endgroup$
– user636268
Jan 21 at 17:49
1
$begingroup$
@Md.ShamimAkhtar You could give it some thought yourself. You might learn something from it.
$endgroup$
– Servaes
Jan 21 at 18:09
1
$begingroup$
A bound is what you asked for.
$endgroup$
– Ross Millikan
Jan 21 at 18:15
1
$begingroup$
Note: saves time to note that your number can't be divisible by both $2$ and $5$ (else it would end in $0$ and the product of its digits would be $0$).
$endgroup$
– lulu
Jan 21 at 19:04
1
$begingroup$
I don't think this is a reasonable contest problem unless you can reduce the search space much more. I don't know how and the Wikipedia article only cites the same upper limit I found.
$endgroup$
– Ross Millikan
Jan 22 at 2:17
|
show 2 more comments
$begingroup$
So whats next? You just created the bound, but this is just a mere observation
$endgroup$
– user636268
Jan 21 at 17:49
1
$begingroup$
@Md.ShamimAkhtar You could give it some thought yourself. You might learn something from it.
$endgroup$
– Servaes
Jan 21 at 18:09
1
$begingroup$
A bound is what you asked for.
$endgroup$
– Ross Millikan
Jan 21 at 18:15
1
$begingroup$
Note: saves time to note that your number can't be divisible by both $2$ and $5$ (else it would end in $0$ and the product of its digits would be $0$).
$endgroup$
– lulu
Jan 21 at 19:04
1
$begingroup$
I don't think this is a reasonable contest problem unless you can reduce the search space much more. I don't know how and the Wikipedia article only cites the same upper limit I found.
$endgroup$
– Ross Millikan
Jan 22 at 2:17
$begingroup$
So whats next? You just created the bound, but this is just a mere observation
$endgroup$
– user636268
Jan 21 at 17:49
$begingroup$
So whats next? You just created the bound, but this is just a mere observation
$endgroup$
– user636268
Jan 21 at 17:49
1
1
$begingroup$
@Md.ShamimAkhtar You could give it some thought yourself. You might learn something from it.
$endgroup$
– Servaes
Jan 21 at 18:09
$begingroup$
@Md.ShamimAkhtar You could give it some thought yourself. You might learn something from it.
$endgroup$
– Servaes
Jan 21 at 18:09
1
1
$begingroup$
A bound is what you asked for.
$endgroup$
– Ross Millikan
Jan 21 at 18:15
$begingroup$
A bound is what you asked for.
$endgroup$
– Ross Millikan
Jan 21 at 18:15
1
1
$begingroup$
Note: saves time to note that your number can't be divisible by both $2$ and $5$ (else it would end in $0$ and the product of its digits would be $0$).
$endgroup$
– lulu
Jan 21 at 19:04
$begingroup$
Note: saves time to note that your number can't be divisible by both $2$ and $5$ (else it would end in $0$ and the product of its digits would be $0$).
$endgroup$
– lulu
Jan 21 at 19:04
1
1
$begingroup$
I don't think this is a reasonable contest problem unless you can reduce the search space much more. I don't know how and the Wikipedia article only cites the same upper limit I found.
$endgroup$
– Ross Millikan
Jan 22 at 2:17
$begingroup$
I don't think this is a reasonable contest problem unless you can reduce the search space much more. I don't know how and the Wikipedia article only cites the same upper limit I found.
$endgroup$
– Ross Millikan
Jan 22 at 2:17
|
show 2 more comments
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1
$begingroup$
Did you search the site this time before posting?
$endgroup$
– Dietrich Burde
Jan 21 at 17:26
$begingroup$
What do you mean?
$endgroup$
– user636268
Jan 21 at 17:26
$begingroup$
I mean, whether your question is new here or already has answers, i.e., is a duplicate?
$endgroup$
– Dietrich Burde
Jan 21 at 17:27
6
$begingroup$
There's actually a name for this sort of number! Let $n$ be a number such that $$n = text{(product of n's digits)} times text{(sum of n's digits)}$$ Then $n$ is what we call a "sum-product number." (en.wikipedia.org/wiki/Sum-product_number) In base $10$, there are only four such numbers, only two of which are nontrivial - 135 and 144. It has been proven that the number of such numbers in any given base is also finite. The Wikipedia article linked will be a good starting point for looking into this topic. I don't know much offhand about proving stuff of this sort though.
$endgroup$
– Eevee Trainer
Jan 21 at 17:55
2
$begingroup$
"But i assume this question will not have any such duplicates" No. As a rule of thumb, every idea you have is unoriginal.
$endgroup$
– Servaes
Jan 21 at 18:07