Can a formula that is tailor-made to incorporate answers be correct?
Sample problem:
Find an equation $theta(n)$ for which $theta(n)=left{ begin{array} &0, text{when } nin text{Composed} \ n, text{when } nin text{Prime} end{array} right.$
This problem is from the International Youth Math Challenge $2018$ and since they do not return marked sheets, I am unsure if my solution was correct.
My final answer was: $$theta (n)=n-ncdot text{sgn} left(prod_{i=1}^{infty} |n-p_i|right)$$ where $p_i$ is the $i^{text{th}}$ prime number. This is all I could come up with and to be honest, I am not too happy with it, because I feel like I have basically chosen something that will only give the answer I want. Is this solution correct, mathematically? Is there a better solution?
Note: $text {sgn}(n)$ is the $text{sign}$ or $text{signum}$ function and $$text{sgn}(n)=left{ begin{array} &-1; nlt 0\ 0; n=0\ 1; ngt 0 end{array} right.$$
prime-numbers contest-math
add a comment |
Sample problem:
Find an equation $theta(n)$ for which $theta(n)=left{ begin{array} &0, text{when } nin text{Composed} \ n, text{when } nin text{Prime} end{array} right.$
This problem is from the International Youth Math Challenge $2018$ and since they do not return marked sheets, I am unsure if my solution was correct.
My final answer was: $$theta (n)=n-ncdot text{sgn} left(prod_{i=1}^{infty} |n-p_i|right)$$ where $p_i$ is the $i^{text{th}}$ prime number. This is all I could come up with and to be honest, I am not too happy with it, because I feel like I have basically chosen something that will only give the answer I want. Is this solution correct, mathematically? Is there a better solution?
Note: $text {sgn}(n)$ is the $text{sign}$ or $text{signum}$ function and $$text{sgn}(n)=left{ begin{array} &-1; nlt 0\ 0; n=0\ 1; ngt 0 end{array} right.$$
prime-numbers contest-math
add a comment |
Sample problem:
Find an equation $theta(n)$ for which $theta(n)=left{ begin{array} &0, text{when } nin text{Composed} \ n, text{when } nin text{Prime} end{array} right.$
This problem is from the International Youth Math Challenge $2018$ and since they do not return marked sheets, I am unsure if my solution was correct.
My final answer was: $$theta (n)=n-ncdot text{sgn} left(prod_{i=1}^{infty} |n-p_i|right)$$ where $p_i$ is the $i^{text{th}}$ prime number. This is all I could come up with and to be honest, I am not too happy with it, because I feel like I have basically chosen something that will only give the answer I want. Is this solution correct, mathematically? Is there a better solution?
Note: $text {sgn}(n)$ is the $text{sign}$ or $text{signum}$ function and $$text{sgn}(n)=left{ begin{array} &-1; nlt 0\ 0; n=0\ 1; ngt 0 end{array} right.$$
prime-numbers contest-math
Sample problem:
Find an equation $theta(n)$ for which $theta(n)=left{ begin{array} &0, text{when } nin text{Composed} \ n, text{when } nin text{Prime} end{array} right.$
This problem is from the International Youth Math Challenge $2018$ and since they do not return marked sheets, I am unsure if my solution was correct.
My final answer was: $$theta (n)=n-ncdot text{sgn} left(prod_{i=1}^{infty} |n-p_i|right)$$ where $p_i$ is the $i^{text{th}}$ prime number. This is all I could come up with and to be honest, I am not too happy with it, because I feel like I have basically chosen something that will only give the answer I want. Is this solution correct, mathematically? Is there a better solution?
Note: $text {sgn}(n)$ is the $text{sign}$ or $text{signum}$ function and $$text{sgn}(n)=left{ begin{array} &-1; nlt 0\ 0; n=0\ 1; ngt 0 end{array} right.$$
prime-numbers contest-math
prime-numbers contest-math
edited 1 hour ago
asked 1 hour ago
Mohammad Zuhair Khan
1,4232525
1,4232525
add a comment |
add a comment |
3 Answers
3
active
oldest
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I'm not sure what kind of functions are allowed but here is a similar one (might be equivalent after some small changes), the differences being it's finite and doesn't require ability to select primes:
For any positive integer $p$, define this function
$$
f(n,p):= leftlceil frac{n-plfloor frac{n}{p}rfloor}{n} rightrceil
$$
If $p$ divides $n$ then $f(n,p)=0$, otherwise $n-plfloor n/prfloorneq 0$ so $f(n,p)=1$.
You can then use this to make the following:
$$
theta(n):= n - nprod_{p=2}^{n-1}f(n,p)
$$
If $n$ is composite then one of the $p$'s will make the product $0$ and hence $theta(n)=n$. Otherwise $n$ is prime and the product is $1$, giving $theta(n)=0$.
add a comment |
Good try; but there's something that needs fixing. When $n$ is composite, the product diverges to $infty$; so you should define $text {sgn} (infty) = 1$.
Thanks for the advice. But, unless I am mistaken, $+infty gt 0?$
– Mohammad Zuhair Khan
1 hour ago
1
@MohammadZuhairKhan Yes; just personally, I would mention that case, since often people assume that youre working in $mathbb{R}$ instead of the extended number system. I would just put it to make sure there's no chance of getting points off.
– Ovi
1 hour ago
Oh okay thank you. Ofcourse, as it is done already, I can not (and would not) update my answer. But is there any other solution for this problem? I have a suspicion that this might be an open problem, or atleast a sensible version of it will.
– Mohammad Zuhair Khan
1 hour ago
@MohammadZuhairKhan I am thinking about it
– Ovi
1 hour ago
add a comment |
Instead of the sign function, you could potentially use the Kronecker delta, which is defined as
$$delta_{mn}=begin{cases}
1 & text{if }n=m\
0 & text{if }nneq m
end{cases}$$
It basically compares two numbers and gives $1$ if there is a match and $0$ otherwise. By summing over such Kronecker delta's, you could build:
$$theta(n)=nsum_{i=1}^{infty}delta_{np_i}$$
(where $p_i$ is the $i$-th prime number). However, I'm wondering if this would be accepted because it kind of bypasses the question of "checking if $n$ is prime". Both our formulas are just a nice rewording of the "text-form" formula given in the question, so I'm not certain this was the kind of answer that was expected.
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
I'm not sure what kind of functions are allowed but here is a similar one (might be equivalent after some small changes), the differences being it's finite and doesn't require ability to select primes:
For any positive integer $p$, define this function
$$
f(n,p):= leftlceil frac{n-plfloor frac{n}{p}rfloor}{n} rightrceil
$$
If $p$ divides $n$ then $f(n,p)=0$, otherwise $n-plfloor n/prfloorneq 0$ so $f(n,p)=1$.
You can then use this to make the following:
$$
theta(n):= n - nprod_{p=2}^{n-1}f(n,p)
$$
If $n$ is composite then one of the $p$'s will make the product $0$ and hence $theta(n)=n$. Otherwise $n$ is prime and the product is $1$, giving $theta(n)=0$.
add a comment |
I'm not sure what kind of functions are allowed but here is a similar one (might be equivalent after some small changes), the differences being it's finite and doesn't require ability to select primes:
For any positive integer $p$, define this function
$$
f(n,p):= leftlceil frac{n-plfloor frac{n}{p}rfloor}{n} rightrceil
$$
If $p$ divides $n$ then $f(n,p)=0$, otherwise $n-plfloor n/prfloorneq 0$ so $f(n,p)=1$.
You can then use this to make the following:
$$
theta(n):= n - nprod_{p=2}^{n-1}f(n,p)
$$
If $n$ is composite then one of the $p$'s will make the product $0$ and hence $theta(n)=n$. Otherwise $n$ is prime and the product is $1$, giving $theta(n)=0$.
add a comment |
I'm not sure what kind of functions are allowed but here is a similar one (might be equivalent after some small changes), the differences being it's finite and doesn't require ability to select primes:
For any positive integer $p$, define this function
$$
f(n,p):= leftlceil frac{n-plfloor frac{n}{p}rfloor}{n} rightrceil
$$
If $p$ divides $n$ then $f(n,p)=0$, otherwise $n-plfloor n/prfloorneq 0$ so $f(n,p)=1$.
You can then use this to make the following:
$$
theta(n):= n - nprod_{p=2}^{n-1}f(n,p)
$$
If $n$ is composite then one of the $p$'s will make the product $0$ and hence $theta(n)=n$. Otherwise $n$ is prime and the product is $1$, giving $theta(n)=0$.
I'm not sure what kind of functions are allowed but here is a similar one (might be equivalent after some small changes), the differences being it's finite and doesn't require ability to select primes:
For any positive integer $p$, define this function
$$
f(n,p):= leftlceil frac{n-plfloor frac{n}{p}rfloor}{n} rightrceil
$$
If $p$ divides $n$ then $f(n,p)=0$, otherwise $n-plfloor n/prfloorneq 0$ so $f(n,p)=1$.
You can then use this to make the following:
$$
theta(n):= n - nprod_{p=2}^{n-1}f(n,p)
$$
If $n$ is composite then one of the $p$'s will make the product $0$ and hence $theta(n)=n$. Otherwise $n$ is prime and the product is $1$, giving $theta(n)=0$.
answered 28 mins ago
Yong Hao Ng
3,1891220
3,1891220
add a comment |
add a comment |
Good try; but there's something that needs fixing. When $n$ is composite, the product diverges to $infty$; so you should define $text {sgn} (infty) = 1$.
Thanks for the advice. But, unless I am mistaken, $+infty gt 0?$
– Mohammad Zuhair Khan
1 hour ago
1
@MohammadZuhairKhan Yes; just personally, I would mention that case, since often people assume that youre working in $mathbb{R}$ instead of the extended number system. I would just put it to make sure there's no chance of getting points off.
– Ovi
1 hour ago
Oh okay thank you. Ofcourse, as it is done already, I can not (and would not) update my answer. But is there any other solution for this problem? I have a suspicion that this might be an open problem, or atleast a sensible version of it will.
– Mohammad Zuhair Khan
1 hour ago
@MohammadZuhairKhan I am thinking about it
– Ovi
1 hour ago
add a comment |
Good try; but there's something that needs fixing. When $n$ is composite, the product diverges to $infty$; so you should define $text {sgn} (infty) = 1$.
Thanks for the advice. But, unless I am mistaken, $+infty gt 0?$
– Mohammad Zuhair Khan
1 hour ago
1
@MohammadZuhairKhan Yes; just personally, I would mention that case, since often people assume that youre working in $mathbb{R}$ instead of the extended number system. I would just put it to make sure there's no chance of getting points off.
– Ovi
1 hour ago
Oh okay thank you. Ofcourse, as it is done already, I can not (and would not) update my answer. But is there any other solution for this problem? I have a suspicion that this might be an open problem, or atleast a sensible version of it will.
– Mohammad Zuhair Khan
1 hour ago
@MohammadZuhairKhan I am thinking about it
– Ovi
1 hour ago
add a comment |
Good try; but there's something that needs fixing. When $n$ is composite, the product diverges to $infty$; so you should define $text {sgn} (infty) = 1$.
Good try; but there's something that needs fixing. When $n$ is composite, the product diverges to $infty$; so you should define $text {sgn} (infty) = 1$.
answered 1 hour ago
Ovi
12.2k1038109
12.2k1038109
Thanks for the advice. But, unless I am mistaken, $+infty gt 0?$
– Mohammad Zuhair Khan
1 hour ago
1
@MohammadZuhairKhan Yes; just personally, I would mention that case, since often people assume that youre working in $mathbb{R}$ instead of the extended number system. I would just put it to make sure there's no chance of getting points off.
– Ovi
1 hour ago
Oh okay thank you. Ofcourse, as it is done already, I can not (and would not) update my answer. But is there any other solution for this problem? I have a suspicion that this might be an open problem, or atleast a sensible version of it will.
– Mohammad Zuhair Khan
1 hour ago
@MohammadZuhairKhan I am thinking about it
– Ovi
1 hour ago
add a comment |
Thanks for the advice. But, unless I am mistaken, $+infty gt 0?$
– Mohammad Zuhair Khan
1 hour ago
1
@MohammadZuhairKhan Yes; just personally, I would mention that case, since often people assume that youre working in $mathbb{R}$ instead of the extended number system. I would just put it to make sure there's no chance of getting points off.
– Ovi
1 hour ago
Oh okay thank you. Ofcourse, as it is done already, I can not (and would not) update my answer. But is there any other solution for this problem? I have a suspicion that this might be an open problem, or atleast a sensible version of it will.
– Mohammad Zuhair Khan
1 hour ago
@MohammadZuhairKhan I am thinking about it
– Ovi
1 hour ago
Thanks for the advice. But, unless I am mistaken, $+infty gt 0?$
– Mohammad Zuhair Khan
1 hour ago
Thanks for the advice. But, unless I am mistaken, $+infty gt 0?$
– Mohammad Zuhair Khan
1 hour ago
1
1
@MohammadZuhairKhan Yes; just personally, I would mention that case, since often people assume that youre working in $mathbb{R}$ instead of the extended number system. I would just put it to make sure there's no chance of getting points off.
– Ovi
1 hour ago
@MohammadZuhairKhan Yes; just personally, I would mention that case, since often people assume that youre working in $mathbb{R}$ instead of the extended number system. I would just put it to make sure there's no chance of getting points off.
– Ovi
1 hour ago
Oh okay thank you. Ofcourse, as it is done already, I can not (and would not) update my answer. But is there any other solution for this problem? I have a suspicion that this might be an open problem, or atleast a sensible version of it will.
– Mohammad Zuhair Khan
1 hour ago
Oh okay thank you. Ofcourse, as it is done already, I can not (and would not) update my answer. But is there any other solution for this problem? I have a suspicion that this might be an open problem, or atleast a sensible version of it will.
– Mohammad Zuhair Khan
1 hour ago
@MohammadZuhairKhan I am thinking about it
– Ovi
1 hour ago
@MohammadZuhairKhan I am thinking about it
– Ovi
1 hour ago
add a comment |
Instead of the sign function, you could potentially use the Kronecker delta, which is defined as
$$delta_{mn}=begin{cases}
1 & text{if }n=m\
0 & text{if }nneq m
end{cases}$$
It basically compares two numbers and gives $1$ if there is a match and $0$ otherwise. By summing over such Kronecker delta's, you could build:
$$theta(n)=nsum_{i=1}^{infty}delta_{np_i}$$
(where $p_i$ is the $i$-th prime number). However, I'm wondering if this would be accepted because it kind of bypasses the question of "checking if $n$ is prime". Both our formulas are just a nice rewording of the "text-form" formula given in the question, so I'm not certain this was the kind of answer that was expected.
add a comment |
Instead of the sign function, you could potentially use the Kronecker delta, which is defined as
$$delta_{mn}=begin{cases}
1 & text{if }n=m\
0 & text{if }nneq m
end{cases}$$
It basically compares two numbers and gives $1$ if there is a match and $0$ otherwise. By summing over such Kronecker delta's, you could build:
$$theta(n)=nsum_{i=1}^{infty}delta_{np_i}$$
(where $p_i$ is the $i$-th prime number). However, I'm wondering if this would be accepted because it kind of bypasses the question of "checking if $n$ is prime". Both our formulas are just a nice rewording of the "text-form" formula given in the question, so I'm not certain this was the kind of answer that was expected.
add a comment |
Instead of the sign function, you could potentially use the Kronecker delta, which is defined as
$$delta_{mn}=begin{cases}
1 & text{if }n=m\
0 & text{if }nneq m
end{cases}$$
It basically compares two numbers and gives $1$ if there is a match and $0$ otherwise. By summing over such Kronecker delta's, you could build:
$$theta(n)=nsum_{i=1}^{infty}delta_{np_i}$$
(where $p_i$ is the $i$-th prime number). However, I'm wondering if this would be accepted because it kind of bypasses the question of "checking if $n$ is prime". Both our formulas are just a nice rewording of the "text-form" formula given in the question, so I'm not certain this was the kind of answer that was expected.
Instead of the sign function, you could potentially use the Kronecker delta, which is defined as
$$delta_{mn}=begin{cases}
1 & text{if }n=m\
0 & text{if }nneq m
end{cases}$$
It basically compares two numbers and gives $1$ if there is a match and $0$ otherwise. By summing over such Kronecker delta's, you could build:
$$theta(n)=nsum_{i=1}^{infty}delta_{np_i}$$
(where $p_i$ is the $i$-th prime number). However, I'm wondering if this would be accepted because it kind of bypasses the question of "checking if $n$ is prime". Both our formulas are just a nice rewording of the "text-form" formula given in the question, so I'm not certain this was the kind of answer that was expected.
answered 37 mins ago
orion2112
436210
436210
add a comment |
add a comment |
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