Mixing
Proving $10240…02401$ composite
$begingroup$
I got this question recently, and have been unable to solve it.
Prove that $1024underbrace{00 ldotsldots 00}_{2014 text{ times}}2401$ is composite.
I have two different ways in mind.
First is $7^4+400(2^2cdot10^{504})^4$, which looks like Sophie Germain, but I'm not sure what to do with the $400$. Another thought is that this is almost a palindrome, with the order of just two digits interchanged. I'm not sure where to go from there, and if it'd provide any results, but it seems interesting nonetheless.
Please help.
elementary-number-theory contest-math arithmetic
asked Jul 27 '18 at 9:14
MalayTheDynamoMalayTheDynamo
1,627934
$endgroup$
|
show 16 more comments
$begingroup$
I got this question recently, and have been unable to solve it.
Prove that $1024underbrace{00 ldotsldots 00}_{2014 text{ times}}2401$ is composite.
I have two different ways in mind.
First is $7^4+400(2^2cdot10^{504})^4$, which looks like Sophie Germain, but I'm not sure what to do with the $400$. Another thought is that this is almost a palindrome, with the order of just two digits interchanged. I'm not sure where to go from there, and if it'd provide any results, but it seems interesting nonetheless.
Please help.
elementary-number-theory contest-math arithmetic
asked Jul 27 '18 at 9:14
MalayTheDynamoMalayTheDynamo
1,627934
$endgroup$
3
$begingroup$
What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
$endgroup$
– Surb
Jul 27 '18 at 9:23
1
$begingroup$
It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
$endgroup$
– MalayTheDynamo
Jul 27 '18 at 9:24
2
$begingroup$
Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
$endgroup$
– Mark Bennet
Jul 27 '18 at 9:28
2
$begingroup$
@Servaes The strong Fermat test to base $2$ says it's composite.
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:24
4
$begingroup$
@Servaes Write $n-1 = 2^kcdot m$ with $m$ odd, and then $$a^{n-1} - 1 = (a^m - 1)prod_{kappa = 0}^{k-1}bigl(a^{2^{kappa} m} + 1bigr)$$ where $1 < a < n-1$. Then $n$ is a strong Fermat probable prime if $n$ divides one of the factors of the product, i.e. $a^m equiv 1 pmod{n}$ or there is a $kappa in {0,dotsc, k-1}$ such that $a^{2^{kappa}m} equiv -1pmod{n}$. That's the test Miller-Rabin is composed of. In this case, $a = 2$ shows it's composite. (Actually already the ordinary Fermat test shows that here.)
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:50
|
show 16 more comments
$begingroup$
I got this question recently, and have been unable to solve it.
Prove that $1024underbrace{00 ldotsldots 00}_{2014 text{ times}}2401$ is composite.
I have two different ways in mind.
First is $7^4+400(2^2cdot10^{504})^4$, which looks like Sophie Germain, but I'm not sure what to do with the $400$. Another thought is that this is almost a palindrome, with the order of just two digits interchanged. I'm not sure where to go from there, and if it'd provide any results, but it seems interesting nonetheless.
Please help.
elementary-number-theory contest-math arithmetic
asked Jul 27 '18 at 9:14
MalayTheDynamoMalayTheDynamo
1,627934
$endgroup$
I got this question recently, and have been unable to solve it.
Prove that $1024underbrace{00 ldotsldots 00}_{2014 text{ times}}2401$ is composite.
I have two different ways in mind.
First is $7^4+400(2^2cdot10^{504})^4$, which looks like Sophie Germain, but I'm not sure what to do with the $400$. Another thought is that this is almost a palindrome, with the order of just two digits interchanged. I'm not sure where to go from there, and if it'd provide any results, but it seems interesting nonetheless.
Please help.
elementary-number-theory contest-math arithmetic
elementary-number-theory contest-math arithmetic
asked Jul 27 '18 at 9:14
MalayTheDynamoMalayTheDynamo
1,627934
asked Jul 27 '18 at 9:14
MalayTheDynamoMalayTheDynamo
1,627934
edited Jul 27 '18 at 9:35
MalayTheDynamo
asked Jul 27 '18 at 9:14
MalayTheDynamoMalayTheDynamo
1,627934
asked Jul 27 '18 at 9:14
MalayTheDynamoMalayTheDynamo
1,627934
asked Jul 27 '18 at 9:14
MalayTheDynamoMalayTheDynamo
1,627934
1,627934
3
$begingroup$
What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
$endgroup$
– Surb
Jul 27 '18 at 9:23
1
$begingroup$
It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
$endgroup$
– MalayTheDynamo
Jul 27 '18 at 9:24
2
$begingroup$
Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
$endgroup$
– Mark Bennet
Jul 27 '18 at 9:28
2
$begingroup$
@Servaes The strong Fermat test to base $2$ says it's composite.
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:24
4
$begingroup$
@Servaes Write $n-1 = 2^kcdot m$ with $m$ odd, and then $$a^{n-1} - 1 = (a^m - 1)prod_{kappa = 0}^{k-1}bigl(a^{2^{kappa} m} + 1bigr)$$ where $1 < a < n-1$. Then $n$ is a strong Fermat probable prime if $n$ divides one of the factors of the product, i.e. $a^m equiv 1 pmod{n}$ or there is a $kappa in {0,dotsc, k-1}$ such that $a^{2^{kappa}m} equiv -1pmod{n}$. That's the test Miller-Rabin is composed of. In this case, $a = 2$ shows it's composite. (Actually already the ordinary Fermat test shows that here.)
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:50
|
show 16 more comments
3
$begingroup$
What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
$endgroup$
– Surb
Jul 27 '18 at 9:23
1
$begingroup$
It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
$endgroup$
– MalayTheDynamo
Jul 27 '18 at 9:24
2
$begingroup$
Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
$endgroup$
– Mark Bennet
Jul 27 '18 at 9:28
2
$begingroup$
@Servaes The strong Fermat test to base $2$ says it's composite.
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:24
4
$begingroup$
@Servaes Write $n-1 = 2^kcdot m$ with $m$ odd, and then $$a^{n-1} - 1 = (a^m - 1)prod_{kappa = 0}^{k-1}bigl(a^{2^{kappa} m} + 1bigr)$$ where $1 < a < n-1$. Then $n$ is a strong Fermat probable prime if $n$ divides one of the factors of the product, i.e. $a^m equiv 1 pmod{n}$ or there is a $kappa in {0,dotsc, k-1}$ such that $a^{2^{kappa}m} equiv -1pmod{n}$. That's the test Miller-Rabin is composed of. In this case, $a = 2$ shows it's composite. (Actually already the ordinary Fermat test shows that here.)
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:50
3
3
$begingroup$
What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
$endgroup$
– Surb
Jul 27 '18 at 9:23
$begingroup$
What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
$endgroup$
– Surb
Jul 27 '18 at 9:23
1
1
$begingroup$
It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
$endgroup$
– MalayTheDynamo
Jul 27 '18 at 9:24
$begingroup$
It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
$endgroup$
– MalayTheDynamo
Jul 27 '18 at 9:24
2
2
$begingroup$
Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
$endgroup$
– Mark Bennet
Jul 27 '18 at 9:28
$begingroup$
Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
$endgroup$
– Mark Bennet
Jul 27 '18 at 9:28
2
2
$begingroup$
@Servaes The strong Fermat test to base $2$ says it's composite.
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:24
$begingroup$
@Servaes The strong Fermat test to base $2$ says it's composite.
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:24
4
4
$begingroup$
@Servaes Write $n-1 = 2^kcdot m$ with $m$ odd, and then $$a^{n-1} - 1 = (a^m - 1)prod_{kappa = 0}^{k-1}bigl(a^{2^{kappa} m} + 1bigr)$$ where $1 < a < n-1$. Then $n$ is a strong Fermat probable prime if $n$ divides one of the factors of the product, i.e. $a^m equiv 1 pmod{n}$ or there is a $kappa in {0,dotsc, k-1}$ such that $a^{2^{kappa}m} equiv -1pmod{n}$. That's the test Miller-Rabin is composed of. In this case, $a = 2$ shows it's composite. (Actually already the ordinary Fermat test shows that here.)
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:50
$begingroup$
@Servaes Write $n-1 = 2^kcdot m$ with $m$ odd, and then $$a^{n-1} - 1 = (a^m - 1)prod_{kappa = 0}^{k-1}bigl(a^{2^{kappa} m} + 1bigr)$$ where $1 < a < n-1$. Then $n$ is a strong Fermat probable prime if $n$ divides one of the factors of the product, i.e. $a^m equiv 1 pmod{n}$ or there is a $kappa in {0,dotsc, k-1}$ such that $a^{2^{kappa}m} equiv -1pmod{n}$. That's the test Miller-Rabin is composed of. In this case, $a = 2$ shows it's composite. (Actually already the ordinary Fermat test shows that here.)
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:50
|
show 16 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The method introduced in this modification is for reduction of volume of calculations.
Let's start with a simple example; consider number $N=1024002401=1670477times 613$. We may write:
$N=2^{10}.10^6+24.10^2 +1=2^3.10^2(2^7.10^4+3)+1$
$2^7.10^4+3=1280003 ≡59 mod (613)$
⇒ $N=2^3.10^2(613 k + 59)+1=800times 613times k +800 times 59 +1$
Let $800times 613 =a$ and $800times 59 +1 =b$
⇒ $(a, b)=613$
Now suppose we do not know that $p=613$ and $r=59$ in following relation:
$(a, b)=(800 p, 800 r+1)=p$
$a=800 r +1$ gives infinite numbers which its factors may be candidates for a factor of a number like ($N=2^{alpha}.10^{beta}+3^{gamma}$).
For example giving r numerous values we find that for $r=59$ we get $800times 59+1=47201=7times 11times 613$. Each of theses three factors can be the candidate for p to be tried to find the factor of N. For $N=2^{10}.10^{2018}+7^4$ we may write:
$N=2^{10}.10^{2018}+7^4=2^3.10^2(2^7.10^{2016}+3)+1$
$2^7.10^{2016}+3≡ r mod (p)= k.p +r$
$N=800(k.p+r)+1=800p.k+800r +1$
Now values for $r$ in $(800 r +1)$ give numbers their factors can be checked as a primes factor of N.
answered Feb 1 at 17:38
siroussirous
1,6611514
$endgroup$
add a comment |
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votes
$begingroup$
The method introduced in this modification is for reduction of volume of calculations.
Let's start with a simple example; consider number $N=1024002401=1670477times 613$. We may write:
$N=2^{10}.10^6+24.10^2 +1=2^3.10^2(2^7.10^4+3)+1$
$2^7.10^4+3=1280003 ≡59 mod (613)$
⇒ $N=2^3.10^2(613 k + 59)+1=800times 613times k +800 times 59 +1$
Let $800times 613 =a$ and $800times 59 +1 =b$
⇒ $(a, b)=613$
Now suppose we do not know that $p=613$ and $r=59$ in following relation:
$(a, b)=(800 p, 800 r+1)=p$
$a=800 r +1$ gives infinite numbers which its factors may be candidates for a factor of a number like ($N=2^{alpha}.10^{beta}+3^{gamma}$).
For example giving r numerous values we find that for $r=59$ we get $800times 59+1=47201=7times 11times 613$. Each of theses three factors can be the candidate for p to be tried to find the factor of N. For $N=2^{10}.10^{2018}+7^4$ we may write:
$N=2^{10}.10^{2018}+7^4=2^3.10^2(2^7.10^{2016}+3)+1$
$2^7.10^{2016}+3≡ r mod (p)= k.p +r$
$N=800(k.p+r)+1=800p.k+800r +1$
Now values for $r$ in $(800 r +1)$ give numbers their factors can be checked as a primes factor of N.
answered Feb 1 at 17:38
siroussirous
1,6611514
$endgroup$
add a comment |
$begingroup$
The method introduced in this modification is for reduction of volume of calculations.
Let's start with a simple example; consider number $N=1024002401=1670477times 613$. We may write:
$N=2^{10}.10^6+24.10^2 +1=2^3.10^2(2^7.10^4+3)+1$
$2^7.10^4+3=1280003 ≡59 mod (613)$
⇒ $N=2^3.10^2(613 k + 59)+1=800times 613times k +800 times 59 +1$
Let $800times 613 =a$ and $800times 59 +1 =b$
⇒ $(a, b)=613$
Now suppose we do not know that $p=613$ and $r=59$ in following relation:
$(a, b)=(800 p, 800 r+1)=p$
$a=800 r +1$ gives infinite numbers which its factors may be candidates for a factor of a number like ($N=2^{alpha}.10^{beta}+3^{gamma}$).
For example giving r numerous values we find that for $r=59$ we get $800times 59+1=47201=7times 11times 613$. Each of theses three factors can be the candidate for p to be tried to find the factor of N. For $N=2^{10}.10^{2018}+7^4$ we may write:
$N=2^{10}.10^{2018}+7^4=2^3.10^2(2^7.10^{2016}+3)+1$
$2^7.10^{2016}+3≡ r mod (p)= k.p +r$
$N=800(k.p+r)+1=800p.k+800r +1$
Now values for $r$ in $(800 r +1)$ give numbers their factors can be checked as a primes factor of N.
answered Feb 1 at 17:38
siroussirous
1,6611514
$endgroup$
add a comment |
$begingroup$
The method introduced in this modification is for reduction of volume of calculations.
Let's start with a simple example; consider number $N=1024002401=1670477times 613$. We may write:
$N=2^{10}.10^6+24.10^2 +1=2^3.10^2(2^7.10^4+3)+1$
$2^7.10^4+3=1280003 ≡59 mod (613)$
⇒ $N=2^3.10^2(613 k + 59)+1=800times 613times k +800 times 59 +1$
Let $800times 613 =a$ and $800times 59 +1 =b$
⇒ $(a, b)=613$
Now suppose we do not know that $p=613$ and $r=59$ in following relation:
$(a, b)=(800 p, 800 r+1)=p$
$a=800 r +1$ gives infinite numbers which its factors may be candidates for a factor of a number like ($N=2^{alpha}.10^{beta}+3^{gamma}$).
For example giving r numerous values we find that for $r=59$ we get $800times 59+1=47201=7times 11times 613$. Each of theses three factors can be the candidate for p to be tried to find the factor of N. For $N=2^{10}.10^{2018}+7^4$ we may write:
$N=2^{10}.10^{2018}+7^4=2^3.10^2(2^7.10^{2016}+3)+1$
$2^7.10^{2016}+3≡ r mod (p)= k.p +r$
$N=800(k.p+r)+1=800p.k+800r +1$
Now values for $r$ in $(800 r +1)$ give numbers their factors can be checked as a primes factor of N.
answered Feb 1 at 17:38
siroussirous
1,6611514
$endgroup$
The method introduced in this modification is for reduction of volume of calculations.
Let's start with a simple example; consider number $N=1024002401=1670477times 613$. We may write:
$N=2^{10}.10^6+24.10^2 +1=2^3.10^2(2^7.10^4+3)+1$
$2^7.10^4+3=1280003 ≡59 mod (613)$
⇒ $N=2^3.10^2(613 k + 59)+1=800times 613times k +800 times 59 +1$
Let $800times 613 =a$ and $800times 59 +1 =b$
⇒ $(a, b)=613$
Now suppose we do not know that $p=613$ and $r=59$ in following relation:
$(a, b)=(800 p, 800 r+1)=p$
$a=800 r +1$ gives infinite numbers which its factors may be candidates for a factor of a number like ($N=2^{alpha}.10^{beta}+3^{gamma}$).
For example giving r numerous values we find that for $r=59$ we get $800times 59+1=47201=7times 11times 613$. Each of theses three factors can be the candidate for p to be tried to find the factor of N. For $N=2^{10}.10^{2018}+7^4$ we may write:
$N=2^{10}.10^{2018}+7^4=2^3.10^2(2^7.10^{2016}+3)+1$
$2^7.10^{2016}+3≡ r mod (p)= k.p +r$
$N=800(k.p+r)+1=800p.k+800r +1$
Now values for $r$ in $(800 r +1)$ give numbers their factors can be checked as a primes factor of N.
answered Feb 1 at 17:38
siroussirous
1,6611514
edited Feb 2 at 5:24
answered Feb 1 at 17:38
siroussirous
1,6611514
answered Feb 1 at 17:38
siroussirous
1,6611514
answered Feb 1 at 17:38
siroussirous
1,6611514
1,6611514
add a comment |
add a comment |
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3
$begingroup$
What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
$endgroup$
– Surb
Jul 27 '18 at 9:23
1
$begingroup$
It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
$endgroup$
– MalayTheDynamo
Jul 27 '18 at 9:24
2
$begingroup$
Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
$endgroup$
– Mark Bennet
Jul 27 '18 at 9:28
2
$begingroup$
@Servaes The strong Fermat test to base $2$ says it's composite.
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:24
4
$begingroup$
@Servaes Write $n-1 = 2^kcdot m$ with $m$ odd, and then $$a^{n-1} - 1 = (a^m - 1)prod_{kappa = 0}^{k-1}bigl(a^{2^{kappa} m} + 1bigr)$$ where $1 < a < n-1$. Then $n$ is a strong Fermat probable prime if $n$ divides one of the factors of the product, i.e. $a^m equiv 1 pmod{n}$ or there is a $kappa in {0,dotsc, k-1}$ such that $a^{2^{kappa}m} equiv -1pmod{n}$. That's the test Miller-Rabin is composed of. In this case, $a = 2$ shows it's composite. (Actually already the ordinary Fermat test shows that here.)
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:50