Is there any other number that has similar properties as $21$?
$begingroup$
It's my observation.
Let
$$n=p_1×p_2×p_3×dots×p_r$$
where $p_i$ are prime factors and
$f$ and $g$ are the functions
$$f(n)=1+2+dots+n$$
And
$$g(n)=p_1+p_2+dots+p_r$$
If we put $n=21$ then
$$g(f(21))=g(231)=21.$$
I checked it upto $n=10000$, I did not find another number with this property $g(f(n))=n$.
Can we prove that other such numbers do not exist?
number-theory elementary-number-theory prime-factorization
edited yesterday
Asaf Karagila♦
305k33435766
asked yesterday
Pruthviraj HajariPruthviraj Hajari
16517
New contributor
Pruthviraj Hajari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
|
show 1 more comment
$begingroup$
It's my observation.
Let
$$n=p_1×p_2×p_3×dots×p_r$$
where $p_i$ are prime factors and
$f$ and $g$ are the functions
$$f(n)=1+2+dots+n$$
And
$$g(n)=p_1+p_2+dots+p_r$$
If we put $n=21$ then
$$g(f(21))=g(231)=21.$$
I checked it upto $n=10000$, I did not find another number with this property $g(f(n))=n$.
Can we prove that other such numbers do not exist?
number-theory elementary-number-theory prime-factorization
edited yesterday
Asaf Karagila♦
305k33435766
asked yesterday
Pruthviraj HajariPruthviraj Hajari
16517
New contributor
Pruthviraj Hajari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
To clarify, is $g(12)=5$ or $=7$?
$endgroup$
– Hagen von Eitzen
yesterday
$begingroup$
12=3*2*2 then g(12)=2+2+3=7
$endgroup$
– Pruthviraj Hajari
yesterday
2
$begingroup$
@PruthvirajHajari Did you come up with this problem by yourself? If so, please state that in your question and include the program code used to run it.
$endgroup$
– TheSimpliFire
yesterday
$begingroup$
I encourage you to accept my answer by clicking the tick you see beside it.
$endgroup$
– Parcly Taxel
yesterday
1
$begingroup$
@PruthvirajHajari sure it is.
$endgroup$
– Parcly Taxel
yesterday
|
show 1 more comment
$begingroup$
It's my observation.
Let
$$n=p_1×p_2×p_3×dots×p_r$$
where $p_i$ are prime factors and
$f$ and $g$ are the functions
$$f(n)=1+2+dots+n$$
And
$$g(n)=p_1+p_2+dots+p_r$$
If we put $n=21$ then
$$g(f(21))=g(231)=21.$$
I checked it upto $n=10000$, I did not find another number with this property $g(f(n))=n$.
Can we prove that other such numbers do not exist?
number-theory elementary-number-theory prime-factorization
edited yesterday
Asaf Karagila♦
305k33435766
asked yesterday
Pruthviraj HajariPruthviraj Hajari
16517
New contributor
Pruthviraj Hajari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
It's my observation.
Let
$$n=p_1×p_2×p_3×dots×p_r$$
where $p_i$ are prime factors and
$f$ and $g$ are the functions
$$f(n)=1+2+dots+n$$
And
$$g(n)=p_1+p_2+dots+p_r$$
If we put $n=21$ then
$$g(f(21))=g(231)=21.$$
I checked it upto $n=10000$, I did not find another number with this property $g(f(n))=n$.
Can we prove that other such numbers do not exist?
number-theory elementary-number-theory prime-factorization
number-theory elementary-number-theory prime-factorization
edited yesterday
Asaf Karagila♦
305k33435766
asked yesterday
Pruthviraj HajariPruthviraj Hajari
16517
New contributor
Pruthviraj Hajari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Asaf Karagila♦
305k33435766
asked yesterday
Pruthviraj HajariPruthviraj Hajari
16517
New contributor
Pruthviraj Hajari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Asaf Karagila♦
305k33435766
edited yesterday
Asaf Karagila♦
305k33435766
edited yesterday
Asaf Karagila♦
305k33435766
305k33435766
asked yesterday
Pruthviraj HajariPruthviraj Hajari
16517
New contributor
Pruthviraj Hajari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Pruthviraj HajariPruthviraj Hajari
16517
asked yesterday
Pruthviraj HajariPruthviraj Hajari
16517
16517
New contributor
Pruthviraj Hajari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Pruthviraj Hajari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Pruthviraj Hajari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
To clarify, is $g(12)=5$ or $=7$?
$endgroup$
– Hagen von Eitzen
yesterday
$begingroup$
12=3*2*2 then g(12)=2+2+3=7
$endgroup$
– Pruthviraj Hajari
yesterday
2
$begingroup$
@PruthvirajHajari Did you come up with this problem by yourself? If so, please state that in your question and include the program code used to run it.
$endgroup$
– TheSimpliFire
yesterday
$begingroup$
I encourage you to accept my answer by clicking the tick you see beside it.
$endgroup$
– Parcly Taxel
yesterday
1
$begingroup$
@PruthvirajHajari sure it is.
$endgroup$
– Parcly Taxel
yesterday
|
show 1 more comment
$begingroup$
To clarify, is $g(12)=5$ or $=7$?
$endgroup$
– Hagen von Eitzen
yesterday
$begingroup$
12=3*2*2 then g(12)=2+2+3=7
$endgroup$
– Pruthviraj Hajari
yesterday
2
$begingroup$
@PruthvirajHajari Did you come up with this problem by yourself? If so, please state that in your question and include the program code used to run it.
$endgroup$
– TheSimpliFire
yesterday
$begingroup$
I encourage you to accept my answer by clicking the tick you see beside it.
$endgroup$
– Parcly Taxel
yesterday
1
$begingroup$
@PruthvirajHajari sure it is.
$endgroup$
– Parcly Taxel
yesterday
$begingroup$
To clarify, is $g(12)=5$ or $=7$?
$endgroup$
– Hagen von Eitzen
yesterday
$begingroup$
To clarify, is $g(12)=5$ or $=7$?
$endgroup$
– Hagen von Eitzen
yesterday
$begingroup$
12=3*2*2 then g(12)=2+2+3=7
$endgroup$
– Pruthviraj Hajari
yesterday
$begingroup$
12=3*2*2 then g(12)=2+2+3=7
$endgroup$
– Pruthviraj Hajari
yesterday
2
2
$begingroup$
@PruthvirajHajari Did you come up with this problem by yourself? If so, please state that in your question and include the program code used to run it.
$endgroup$
– TheSimpliFire
yesterday
$begingroup$
@PruthvirajHajari Did you come up with this problem by yourself? If so, please state that in your question and include the program code used to run it.
$endgroup$
– TheSimpliFire
yesterday
$begingroup$
I encourage you to accept my answer by clicking the tick you see beside it.
$endgroup$
– Parcly Taxel
yesterday
$begingroup$
I encourage you to accept my answer by clicking the tick you see beside it.
$endgroup$
– Parcly Taxel
yesterday
1
1
$begingroup$
@PruthvirajHajari sure it is.
$endgroup$
– Parcly Taxel
yesterday
$begingroup$
@PruthvirajHajari sure it is.
$endgroup$
– Parcly Taxel
yesterday
|
show 1 more comment
1 Answer
1
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$begingroup$
This is a very interesting question…
$newcommand{sopfr}{operatorname{sopfr}}$
$f(n)=frac{n(n+1)}2$ and $g(n)=sopfr(n)$, the sum of prime factors of $n$ with repeats (OEIS A001414). We want $n$ such that $g(f(n))=n$ or
$$sopfrleft(frac{n(n+1)}2right)=ntag1$$
which can be split into two cases due to the property $sopfr(ab)=sopfr(a)+sopfr(b)$.
- If $n$ is even, then $sopfrleft(frac n2right)+sopfr(n+1)=n$. We know that $sopfr(n)le n$, so $sopfrleft(frac n2right)lefrac n2$ and consequently $sopfr(n+1)gefrac n2$. Either $n+1$ is a prime, in which case the LHS of $(1)$ is greater than $n$ and so the equality cannot hold, or $n+1$ is odd composite and so has a least prime factor at least 3*, yielding $sopfr(n+1)le3+frac{n+1}3$ and thus
$$frac n2le3+frac{n+1}3$$
which is only true for $nle20$. Checking those $n$ reveals no solutions to $(1)$. - If $n$ is odd, the reasoning is similar: $sopfrleft(frac{n+1}2right)+sopfr(n)=n$, where $sopfrleft(frac{n+1}2right)lefrac{n+1}2$ and so $sopfr(n)gefrac{n-1}2$. Since $n$ is odd, either it is prime and the LHS of $(1)$ is greater than $n$, or it has a least prime factor at least 3* and $sopfr(n)le3+frac n3$, giving
$$frac{n-1}2le3+frac n3$$
which only holds for $nle21$. 21 is the solution to $(1)$ pointed out in the original question; we have just shown it is the only one.
*Technically we have to repeat the argument for other possible least prime factors $k$ of $n$ or $n+1$ – and the upper bound $N_k$ of the solution to the inequalities in $n$ increases accordingly, each 3 replaced with $k$. However, the least composite number with least prime factor $k$ is $k^2$, and this increases much faster than $N_k$ (which is $simfrac k2$). Indeed, $5^2$ already exceeds $N_5$ for both inequalities.
The method I use above has very strong similarities to the method I used in my most famous answer of all. It is sheer coincidence that 21 is a solution to both the problems I answered.
answered yesterday
Parcly TaxelParcly Taxel
1
$endgroup$
$begingroup$
To elaborate on the footnote: If $n$ is composite and $p$ its minimal prime factor, then $operatorname{sopfr}(n)=p+operatorname{sopfr}(n/p)le p+frac np$ and $xmapsto x+frac nx$ is a decreasing function on $(0,sqrt n)$, hence $n$ odd and composite implies $operatorname{sopfr}(n)le 3+frac n3$.
$endgroup$
– Hagen von Eitzen
yesterday
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
This is a very interesting question…
$newcommand{sopfr}{operatorname{sopfr}}$
$f(n)=frac{n(n+1)}2$ and $g(n)=sopfr(n)$, the sum of prime factors of $n$ with repeats (OEIS A001414). We want $n$ such that $g(f(n))=n$ or
$$sopfrleft(frac{n(n+1)}2right)=ntag1$$
which can be split into two cases due to the property $sopfr(ab)=sopfr(a)+sopfr(b)$.
- If $n$ is even, then $sopfrleft(frac n2right)+sopfr(n+1)=n$. We know that $sopfr(n)le n$, so $sopfrleft(frac n2right)lefrac n2$ and consequently $sopfr(n+1)gefrac n2$. Either $n+1$ is a prime, in which case the LHS of $(1)$ is greater than $n$ and so the equality cannot hold, or $n+1$ is odd composite and so has a least prime factor at least 3*, yielding $sopfr(n+1)le3+frac{n+1}3$ and thus
$$frac n2le3+frac{n+1}3$$
which is only true for $nle20$. Checking those $n$ reveals no solutions to $(1)$. - If $n$ is odd, the reasoning is similar: $sopfrleft(frac{n+1}2right)+sopfr(n)=n$, where $sopfrleft(frac{n+1}2right)lefrac{n+1}2$ and so $sopfr(n)gefrac{n-1}2$. Since $n$ is odd, either it is prime and the LHS of $(1)$ is greater than $n$, or it has a least prime factor at least 3* and $sopfr(n)le3+frac n3$, giving
$$frac{n-1}2le3+frac n3$$
which only holds for $nle21$. 21 is the solution to $(1)$ pointed out in the original question; we have just shown it is the only one.
*Technically we have to repeat the argument for other possible least prime factors $k$ of $n$ or $n+1$ – and the upper bound $N_k$ of the solution to the inequalities in $n$ increases accordingly, each 3 replaced with $k$. However, the least composite number with least prime factor $k$ is $k^2$, and this increases much faster than $N_k$ (which is $simfrac k2$). Indeed, $5^2$ already exceeds $N_5$ for both inequalities.
The method I use above has very strong similarities to the method I used in my most famous answer of all. It is sheer coincidence that 21 is a solution to both the problems I answered.
answered yesterday
Parcly TaxelParcly Taxel
1
$endgroup$
$begingroup$
To elaborate on the footnote: If $n$ is composite and $p$ its minimal prime factor, then $operatorname{sopfr}(n)=p+operatorname{sopfr}(n/p)le p+frac np$ and $xmapsto x+frac nx$ is a decreasing function on $(0,sqrt n)$, hence $n$ odd and composite implies $operatorname{sopfr}(n)le 3+frac n3$.
$endgroup$
– Hagen von Eitzen
yesterday
add a comment |
$begingroup$
This is a very interesting question…
$newcommand{sopfr}{operatorname{sopfr}}$
$f(n)=frac{n(n+1)}2$ and $g(n)=sopfr(n)$, the sum of prime factors of $n$ with repeats (OEIS A001414). We want $n$ such that $g(f(n))=n$ or
$$sopfrleft(frac{n(n+1)}2right)=ntag1$$
which can be split into two cases due to the property $sopfr(ab)=sopfr(a)+sopfr(b)$.
- If $n$ is even, then $sopfrleft(frac n2right)+sopfr(n+1)=n$. We know that $sopfr(n)le n$, so $sopfrleft(frac n2right)lefrac n2$ and consequently $sopfr(n+1)gefrac n2$. Either $n+1$ is a prime, in which case the LHS of $(1)$ is greater than $n$ and so the equality cannot hold, or $n+1$ is odd composite and so has a least prime factor at least 3*, yielding $sopfr(n+1)le3+frac{n+1}3$ and thus
$$frac n2le3+frac{n+1}3$$
which is only true for $nle20$. Checking those $n$ reveals no solutions to $(1)$. - If $n$ is odd, the reasoning is similar: $sopfrleft(frac{n+1}2right)+sopfr(n)=n$, where $sopfrleft(frac{n+1}2right)lefrac{n+1}2$ and so $sopfr(n)gefrac{n-1}2$. Since $n$ is odd, either it is prime and the LHS of $(1)$ is greater than $n$, or it has a least prime factor at least 3* and $sopfr(n)le3+frac n3$, giving
$$frac{n-1}2le3+frac n3$$
which only holds for $nle21$. 21 is the solution to $(1)$ pointed out in the original question; we have just shown it is the only one.
*Technically we have to repeat the argument for other possible least prime factors $k$ of $n$ or $n+1$ – and the upper bound $N_k$ of the solution to the inequalities in $n$ increases accordingly, each 3 replaced with $k$. However, the least composite number with least prime factor $k$ is $k^2$, and this increases much faster than $N_k$ (which is $simfrac k2$). Indeed, $5^2$ already exceeds $N_5$ for both inequalities.
The method I use above has very strong similarities to the method I used in my most famous answer of all. It is sheer coincidence that 21 is a solution to both the problems I answered.
answered yesterday
Parcly TaxelParcly Taxel
1
$endgroup$
$begingroup$
To elaborate on the footnote: If $n$ is composite and $p$ its minimal prime factor, then $operatorname{sopfr}(n)=p+operatorname{sopfr}(n/p)le p+frac np$ and $xmapsto x+frac nx$ is a decreasing function on $(0,sqrt n)$, hence $n$ odd and composite implies $operatorname{sopfr}(n)le 3+frac n3$.
$endgroup$
– Hagen von Eitzen
yesterday
add a comment |
$begingroup$
This is a very interesting question…
$newcommand{sopfr}{operatorname{sopfr}}$
$f(n)=frac{n(n+1)}2$ and $g(n)=sopfr(n)$, the sum of prime factors of $n$ with repeats (OEIS A001414). We want $n$ such that $g(f(n))=n$ or
$$sopfrleft(frac{n(n+1)}2right)=ntag1$$
which can be split into two cases due to the property $sopfr(ab)=sopfr(a)+sopfr(b)$.
- If $n$ is even, then $sopfrleft(frac n2right)+sopfr(n+1)=n$. We know that $sopfr(n)le n$, so $sopfrleft(frac n2right)lefrac n2$ and consequently $sopfr(n+1)gefrac n2$. Either $n+1$ is a prime, in which case the LHS of $(1)$ is greater than $n$ and so the equality cannot hold, or $n+1$ is odd composite and so has a least prime factor at least 3*, yielding $sopfr(n+1)le3+frac{n+1}3$ and thus
$$frac n2le3+frac{n+1}3$$
which is only true for $nle20$. Checking those $n$ reveals no solutions to $(1)$. - If $n$ is odd, the reasoning is similar: $sopfrleft(frac{n+1}2right)+sopfr(n)=n$, where $sopfrleft(frac{n+1}2right)lefrac{n+1}2$ and so $sopfr(n)gefrac{n-1}2$. Since $n$ is odd, either it is prime and the LHS of $(1)$ is greater than $n$, or it has a least prime factor at least 3* and $sopfr(n)le3+frac n3$, giving
$$frac{n-1}2le3+frac n3$$
which only holds for $nle21$. 21 is the solution to $(1)$ pointed out in the original question; we have just shown it is the only one.
*Technically we have to repeat the argument for other possible least prime factors $k$ of $n$ or $n+1$ – and the upper bound $N_k$ of the solution to the inequalities in $n$ increases accordingly, each 3 replaced with $k$. However, the least composite number with least prime factor $k$ is $k^2$, and this increases much faster than $N_k$ (which is $simfrac k2$). Indeed, $5^2$ already exceeds $N_5$ for both inequalities.
The method I use above has very strong similarities to the method I used in my most famous answer of all. It is sheer coincidence that 21 is a solution to both the problems I answered.
answered yesterday
Parcly TaxelParcly Taxel
1
$endgroup$
This is a very interesting question…
$newcommand{sopfr}{operatorname{sopfr}}$
$f(n)=frac{n(n+1)}2$ and $g(n)=sopfr(n)$, the sum of prime factors of $n$ with repeats (OEIS A001414). We want $n$ such that $g(f(n))=n$ or
$$sopfrleft(frac{n(n+1)}2right)=ntag1$$
which can be split into two cases due to the property $sopfr(ab)=sopfr(a)+sopfr(b)$.
- If $n$ is even, then $sopfrleft(frac n2right)+sopfr(n+1)=n$. We know that $sopfr(n)le n$, so $sopfrleft(frac n2right)lefrac n2$ and consequently $sopfr(n+1)gefrac n2$. Either $n+1$ is a prime, in which case the LHS of $(1)$ is greater than $n$ and so the equality cannot hold, or $n+1$ is odd composite and so has a least prime factor at least 3*, yielding $sopfr(n+1)le3+frac{n+1}3$ and thus
$$frac n2le3+frac{n+1}3$$
which is only true for $nle20$. Checking those $n$ reveals no solutions to $(1)$. - If $n$ is odd, the reasoning is similar: $sopfrleft(frac{n+1}2right)+sopfr(n)=n$, where $sopfrleft(frac{n+1}2right)lefrac{n+1}2$ and so $sopfr(n)gefrac{n-1}2$. Since $n$ is odd, either it is prime and the LHS of $(1)$ is greater than $n$, or it has a least prime factor at least 3* and $sopfr(n)le3+frac n3$, giving
$$frac{n-1}2le3+frac n3$$
which only holds for $nle21$. 21 is the solution to $(1)$ pointed out in the original question; we have just shown it is the only one.
*Technically we have to repeat the argument for other possible least prime factors $k$ of $n$ or $n+1$ – and the upper bound $N_k$ of the solution to the inequalities in $n$ increases accordingly, each 3 replaced with $k$. However, the least composite number with least prime factor $k$ is $k^2$, and this increases much faster than $N_k$ (which is $simfrac k2$). Indeed, $5^2$ already exceeds $N_5$ for both inequalities.
The method I use above has very strong similarities to the method I used in my most famous answer of all. It is sheer coincidence that 21 is a solution to both the problems I answered.
answered yesterday
Parcly TaxelParcly Taxel
1
edited yesterday
answered yesterday
Parcly TaxelParcly Taxel
1
answered yesterday
Parcly TaxelParcly Taxel
1
answered yesterday
Parcly TaxelParcly Taxel
1
1
$begingroup$
To elaborate on the footnote: If $n$ is composite and $p$ its minimal prime factor, then $operatorname{sopfr}(n)=p+operatorname{sopfr}(n/p)le p+frac np$ and $xmapsto x+frac nx$ is a decreasing function on $(0,sqrt n)$, hence $n$ odd and composite implies $operatorname{sopfr}(n)le 3+frac n3$.
$endgroup$
– Hagen von Eitzen
yesterday
add a comment |
$begingroup$
To elaborate on the footnote: If $n$ is composite and $p$ its minimal prime factor, then $operatorname{sopfr}(n)=p+operatorname{sopfr}(n/p)le p+frac np$ and $xmapsto x+frac nx$ is a decreasing function on $(0,sqrt n)$, hence $n$ odd and composite implies $operatorname{sopfr}(n)le 3+frac n3$.
$endgroup$
– Hagen von Eitzen
yesterday
$begingroup$
To elaborate on the footnote: If $n$ is composite and $p$ its minimal prime factor, then $operatorname{sopfr}(n)=p+operatorname{sopfr}(n/p)le p+frac np$ and $xmapsto x+frac nx$ is a decreasing function on $(0,sqrt n)$, hence $n$ odd and composite implies $operatorname{sopfr}(n)le 3+frac n3$.
$endgroup$
– Hagen von Eitzen
yesterday
$begingroup$
To elaborate on the footnote: If $n$ is composite and $p$ its minimal prime factor, then $operatorname{sopfr}(n)=p+operatorname{sopfr}(n/p)le p+frac np$ and $xmapsto x+frac nx$ is a decreasing function on $(0,sqrt n)$, hence $n$ odd and composite implies $operatorname{sopfr}(n)le 3+frac n3$.
$endgroup$
– Hagen von Eitzen
yesterday
add a comment |
Pruthviraj Hajari is a new contributor. Be nice, and check out our Code of Conduct.
Pruthviraj Hajari is a new contributor. Be nice, and check out our Code of Conduct.
Pruthviraj Hajari is a new contributor. Be nice, and check out our Code of Conduct.
Pruthviraj Hajari is a new contributor. Be nice, and check out our Code of Conduct.
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To clarify, is $g(12)=5$ or $=7$?
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– Hagen von Eitzen
yesterday
$begingroup$
12=3*2*2 then g(12)=2+2+3=7
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– Pruthviraj Hajari
yesterday
2
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@PruthvirajHajari Did you come up with this problem by yourself? If so, please state that in your question and include the program code used to run it.
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– TheSimpliFire
yesterday
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I encourage you to accept my answer by clicking the tick you see beside it.
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– Parcly Taxel
yesterday
1
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@PruthvirajHajari sure it is.
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– Parcly Taxel
yesterday