Mixing
Find number of ordered pairs
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Find the number of ordered pairs $(p , q)$ such that $p , q $ are both prime numbers less than 50 , and $pq$+1 is divisible by 12
Edit :
What i have done is i have written down all the primes below 50 congruent to modulo 12 .
For example : 11 $ equiv$ -3 $(mod 12)$
number-theory elementary-number-theory prime-numbers
asked Jul 24 at 15:53
mampuuu
304
 |Â
show 1 more comment
up vote
1
down vote
favorite
Find the number of ordered pairs $(p , q)$ such that $p , q $ are both prime numbers less than 50 , and $pq$+1 is divisible by 12
Edit :
What i have done is i have written down all the primes below 50 congruent to modulo 12 .
For example : 11 $ equiv$ -3 $(mod 12)$
number-theory elementary-number-theory prime-numbers
asked Jul 24 at 15:53
mampuuu
304
1
What have you tried in solving the question? Please include that below your question body (as an edit).
– Parcly Taxel
Jul 24 at 15:53
you could do this in excel by hand -- what is the problem exactly?
– gt6989b
Jul 24 at 15:56
@gt6989b Or just by hand, without excel ...
– Peter
Jul 24 at 15:57
@Peter how? There are going to be a lot of combinations
– mampuuu
Jul 24 at 15:59
We have $15$ primes below $50$, making $105$ pairs. Moreover $2$ and $3$ can be omitted because a number of the form $2k+1$ is odd and a number of the form $3k+1$ not disivisble by $3$. So, the number of pairs decreases to $78$. With a bit patience, well feasible by hand.
– Peter
Jul 24 at 16:03
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Find the number of ordered pairs $(p , q)$ such that $p , q $ are both prime numbers less than 50 , and $pq$+1 is divisible by 12
Edit :
What i have done is i have written down all the primes below 50 congruent to modulo 12 .
For example : 11 $ equiv$ -3 $(mod 12)$
number-theory elementary-number-theory prime-numbers
asked Jul 24 at 15:53
mampuuu
304
Find the number of ordered pairs $(p , q)$ such that $p , q $ are both prime numbers less than 50 , and $pq$+1 is divisible by 12
Edit :
What i have done is i have written down all the primes below 50 congruent to modulo 12 .
For example : 11 $ equiv$ -3 $(mod 12)$
number-theory elementary-number-theory prime-numbers
asked Jul 24 at 15:53
mampuuu
304
edited Jul 24 at 15:57
asked Jul 24 at 15:53
mampuuu
304
asked Jul 24 at 15:53
mampuuu
304
asked Jul 24 at 15:53
mampuuu
304
304
1
What have you tried in solving the question? Please include that below your question body (as an edit).
– Parcly Taxel
Jul 24 at 15:53
you could do this in excel by hand -- what is the problem exactly?
– gt6989b
Jul 24 at 15:56
@gt6989b Or just by hand, without excel ...
– Peter
Jul 24 at 15:57
@Peter how? There are going to be a lot of combinations
– mampuuu
Jul 24 at 15:59
We have $15$ primes below $50$, making $105$ pairs. Moreover $2$ and $3$ can be omitted because a number of the form $2k+1$ is odd and a number of the form $3k+1$ not disivisble by $3$. So, the number of pairs decreases to $78$. With a bit patience, well feasible by hand.
– Peter
Jul 24 at 16:03
 |Â
show 1 more comment
1
What have you tried in solving the question? Please include that below your question body (as an edit).
– Parcly Taxel
Jul 24 at 15:53
you could do this in excel by hand -- what is the problem exactly?
– gt6989b
Jul 24 at 15:56
@gt6989b Or just by hand, without excel ...
– Peter
Jul 24 at 15:57
@Peter how? There are going to be a lot of combinations
– mampuuu
Jul 24 at 15:59
We have $15$ primes below $50$, making $105$ pairs. Moreover $2$ and $3$ can be omitted because a number of the form $2k+1$ is odd and a number of the form $3k+1$ not disivisble by $3$. So, the number of pairs decreases to $78$. With a bit patience, well feasible by hand.
– Peter
Jul 24 at 16:03
1
1
What have you tried in solving the question? Please include that below your question body (as an edit).
– Parcly Taxel
Jul 24 at 15:53
What have you tried in solving the question? Please include that below your question body (as an edit).
– Parcly Taxel
Jul 24 at 15:53
you could do this in excel by hand -- what is the problem exactly?
– gt6989b
Jul 24 at 15:56
you could do this in excel by hand -- what is the problem exactly?
– gt6989b
Jul 24 at 15:56
@gt6989b Or just by hand, without excel ...
– Peter
Jul 24 at 15:57
@gt6989b Or just by hand, without excel ...
– Peter
Jul 24 at 15:57
@Peter how? There are going to be a lot of combinations
– mampuuu
Jul 24 at 15:59
@Peter how? There are going to be a lot of combinations
– mampuuu
Jul 24 at 15:59
We have $15$ primes below $50$, making $105$ pairs. Moreover $2$ and $3$ can be omitted because a number of the form $2k+1$ is odd and a number of the form $3k+1$ not disivisble by $3$. So, the number of pairs decreases to $78$. With a bit patience, well feasible by hand.
– Peter
Jul 24 at 16:03
We have $15$ primes below $50$, making $105$ pairs. Moreover $2$ and $3$ can be omitted because a number of the form $2k+1$ is odd and a number of the form $3k+1$ not disivisble by $3$. So, the number of pairs decreases to $78$. With a bit patience, well feasible by hand.
– Peter
Jul 24 at 16:03
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
0
down vote
Hint Note that $p,q$ cannot be 2 or 3. Then $p,q equiv pm 1 pmod6$.
Note that
$$(6k+1)(6n+1) equiv 6(k+n)+1 notequiv -1 pmod12 \
(6k-1)(6n-1) equiv -6(k+n)+1 notequiv -1 pmod12 $$
Therefore, the only posibility left is
$$(6k+1)(6n-1) equiv 6(n-k)-1 pmod12 $$
which is $equiv -1 pmod12$ if and only if $k-n$ is even.
answered Jul 24 at 15:59
N. S.
97.7k5105197
How did you get that $(6k+1)(6n+1)≡6(k+n)+1notequiv−1$ (mod12)?
– mampuuu
Jul 24 at 16:27
@mampuuu Just open the brackets and use the fact that $36 equiv 0 pmod12$. For the last non-equality, note that $6(k+n) equiv 0,6 pmod12$, depending if $k+n$ is even or odd.
– N. S.
Jul 24 at 16:42
add a comment |Â
up vote
0
down vote
Since the numbers are in very small range, I wrote a simple Python script for it. Takes $ < 1 $ second.
def is_prime(x):
if x == 2:
return True
if x % 2 == 0:
return False
for i in range(3, int(x**0.5) + 1, 2):
if x % i == 0:
return False
return True
ans = 0
for p in range(2, 50):
if is_prime(p):
for q in range(2, 50):
if is_prime(q):
ans += (p*q + 1) % 12 == 0
print(ans)
The answer:
44
answered Jul 24 at 19:46
Rahul Goswami
319114
add a comment |Â
up vote
0
down vote
Clearly we need both $p$ and $q$ to be coprime to $12$, so we cannot have $p$ or $q$ equal to either $2$ or $3$.
This means that $p,qin1,5,7,11bmod 12$. We can quickly calculate the products $bmod 12$ across this set:
beginarraycc
p backslash q & 1 & 5 & 7 & 11 \ hline
1 & 1 & 5 & 7 & 11 \
5 & 5 & 1 & 11 & 7 \
7 & 7 & 11 & 1 & 5 \
11 & 11 & 7 & 5 & 1 \
endarray
We are looking for $p,q$ pairs that give $11equiv -1 bmod 12$ in the above table so that $12mid pq+1$. So we would select the pairs from the different residue class sets appropriately, which gives a way to calculate how many options in total.
For example, the set of primes in range $equiv 1 bmod 12$ (call this $S_1$) consists of just $13,37$. This gives us $|S_1|=2$ and similarly $|S_5|=4,$ $|S_7|=4,$ and $|S_11|=3$. Thus we will have $2cdot 3 + 4cdot 4 + 4cdot 4 + 3cdot 2 = 2cdot 6+2cdot 16= fbox44 $ ordered pair solutions.
answered Jul 24 at 16:49
Joffan
31.8k43169
But I guess 1 is not a prime number.
– mampuuu
Jul 24 at 18:48
No it isn't; you're looking (in that case) for the prime numbers that fit $pequiv 1 bmod 12$. Similarly in the other cases you're looking for the sets of prime numbers $equiv 5,7,11 bmod 12$. Example added to answer.
– Joffan
Jul 24 at 18:51
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Hint Note that $p,q$ cannot be 2 or 3. Then $p,q equiv pm 1 pmod6$.
Note that
$$(6k+1)(6n+1) equiv 6(k+n)+1 notequiv -1 pmod12 \
(6k-1)(6n-1) equiv -6(k+n)+1 notequiv -1 pmod12 $$
Therefore, the only posibility left is
$$(6k+1)(6n-1) equiv 6(n-k)-1 pmod12 $$
which is $equiv -1 pmod12$ if and only if $k-n$ is even.
answered Jul 24 at 15:59
N. S.
97.7k5105197
How did you get that $(6k+1)(6n+1)≡6(k+n)+1notequiv−1$ (mod12)?
– mampuuu
Jul 24 at 16:27
@mampuuu Just open the brackets and use the fact that $36 equiv 0 pmod12$. For the last non-equality, note that $6(k+n) equiv 0,6 pmod12$, depending if $k+n$ is even or odd.
– N. S.
Jul 24 at 16:42
add a comment |Â
up vote
0
down vote
Hint Note that $p,q$ cannot be 2 or 3. Then $p,q equiv pm 1 pmod6$.
Note that
$$(6k+1)(6n+1) equiv 6(k+n)+1 notequiv -1 pmod12 \
(6k-1)(6n-1) equiv -6(k+n)+1 notequiv -1 pmod12 $$
Therefore, the only posibility left is
$$(6k+1)(6n-1) equiv 6(n-k)-1 pmod12 $$
which is $equiv -1 pmod12$ if and only if $k-n$ is even.
answered Jul 24 at 15:59
N. S.
97.7k5105197
How did you get that $(6k+1)(6n+1)≡6(k+n)+1notequiv−1$ (mod12)?
– mampuuu
Jul 24 at 16:27
@mampuuu Just open the brackets and use the fact that $36 equiv 0 pmod12$. For the last non-equality, note that $6(k+n) equiv 0,6 pmod12$, depending if $k+n$ is even or odd.
– N. S.
Jul 24 at 16:42
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint Note that $p,q$ cannot be 2 or 3. Then $p,q equiv pm 1 pmod6$.
Note that
$$(6k+1)(6n+1) equiv 6(k+n)+1 notequiv -1 pmod12 \
(6k-1)(6n-1) equiv -6(k+n)+1 notequiv -1 pmod12 $$
Therefore, the only posibility left is
$$(6k+1)(6n-1) equiv 6(n-k)-1 pmod12 $$
which is $equiv -1 pmod12$ if and only if $k-n$ is even.
answered Jul 24 at 15:59
N. S.
97.7k5105197
Hint Note that $p,q$ cannot be 2 or 3. Then $p,q equiv pm 1 pmod6$.
Note that
$$(6k+1)(6n+1) equiv 6(k+n)+1 notequiv -1 pmod12 \
(6k-1)(6n-1) equiv -6(k+n)+1 notequiv -1 pmod12 $$
Therefore, the only posibility left is
$$(6k+1)(6n-1) equiv 6(n-k)-1 pmod12 $$
which is $equiv -1 pmod12$ if and only if $k-n$ is even.
answered Jul 24 at 15:59
N. S.
97.7k5105197
answered Jul 24 at 15:59
N. S.
97.7k5105197
answered Jul 24 at 15:59
N. S.
97.7k5105197
answered Jul 24 at 15:59
N. S.
97.7k5105197
97.7k5105197
How did you get that $(6k+1)(6n+1)≡6(k+n)+1notequiv−1$ (mod12)?
– mampuuu
Jul 24 at 16:27
@mampuuu Just open the brackets and use the fact that $36 equiv 0 pmod12$. For the last non-equality, note that $6(k+n) equiv 0,6 pmod12$, depending if $k+n$ is even or odd.
– N. S.
Jul 24 at 16:42
add a comment |Â
How did you get that $(6k+1)(6n+1)≡6(k+n)+1notequiv−1$ (mod12)?
– mampuuu
Jul 24 at 16:27
@mampuuu Just open the brackets and use the fact that $36 equiv 0 pmod12$. For the last non-equality, note that $6(k+n) equiv 0,6 pmod12$, depending if $k+n$ is even or odd.
– N. S.
Jul 24 at 16:42
How did you get that $(6k+1)(6n+1)≡6(k+n)+1notequiv−1$ (mod12)?
– mampuuu
Jul 24 at 16:27
How did you get that $(6k+1)(6n+1)≡6(k+n)+1notequiv−1$ (mod12)?
– mampuuu
Jul 24 at 16:27
@mampuuu Just open the brackets and use the fact that $36 equiv 0 pmod12$. For the last non-equality, note that $6(k+n) equiv 0,6 pmod12$, depending if $k+n$ is even or odd.
– N. S.
Jul 24 at 16:42
@mampuuu Just open the brackets and use the fact that $36 equiv 0 pmod12$. For the last non-equality, note that $6(k+n) equiv 0,6 pmod12$, depending if $k+n$ is even or odd.
– N. S.
Jul 24 at 16:42
add a comment |Â
up vote
0
down vote
Since the numbers are in very small range, I wrote a simple Python script for it. Takes $ < 1 $ second.
def is_prime(x):
if x == 2:
return True
if x % 2 == 0:
return False
for i in range(3, int(x**0.5) + 1, 2):
if x % i == 0:
return False
return True
ans = 0
for p in range(2, 50):
if is_prime(p):
for q in range(2, 50):
if is_prime(q):
ans += (p*q + 1) % 12 == 0
print(ans)
The answer:
44
answered Jul 24 at 19:46
Rahul Goswami
319114
add a comment |Â
up vote
0
down vote
Since the numbers are in very small range, I wrote a simple Python script for it. Takes $ < 1 $ second.
def is_prime(x):
if x == 2:
return True
if x % 2 == 0:
return False
for i in range(3, int(x**0.5) + 1, 2):
if x % i == 0:
return False
return True
ans = 0
for p in range(2, 50):
if is_prime(p):
for q in range(2, 50):
if is_prime(q):
ans += (p*q + 1) % 12 == 0
print(ans)
The answer:
44
answered Jul 24 at 19:46
Rahul Goswami
319114
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since the numbers are in very small range, I wrote a simple Python script for it. Takes $ < 1 $ second.
def is_prime(x):
if x == 2:
return True
if x % 2 == 0:
return False
for i in range(3, int(x**0.5) + 1, 2):
if x % i == 0:
return False
return True
ans = 0
for p in range(2, 50):
if is_prime(p):
for q in range(2, 50):
if is_prime(q):
ans += (p*q + 1) % 12 == 0
print(ans)
The answer:
44
answered Jul 24 at 19:46
Rahul Goswami
319114
Since the numbers are in very small range, I wrote a simple Python script for it. Takes $ < 1 $ second.
def is_prime(x):
if x == 2:
return True
if x % 2 == 0:
return False
for i in range(3, int(x**0.5) + 1, 2):
if x % i == 0:
return False
return True
ans = 0
for p in range(2, 50):
if is_prime(p):
for q in range(2, 50):
if is_prime(q):
ans += (p*q + 1) % 12 == 0
print(ans)
The answer:
44
answered Jul 24 at 19:46
Rahul Goswami
319114
answered Jul 24 at 19:46
Rahul Goswami
319114
answered Jul 24 at 19:46
Rahul Goswami
319114
answered Jul 24 at 19:46
Rahul Goswami
319114
319114
add a comment |Â
add a comment |Â
up vote
0
down vote
Clearly we need both $p$ and $q$ to be coprime to $12$, so we cannot have $p$ or $q$ equal to either $2$ or $3$.
This means that $p,qin1,5,7,11bmod 12$. We can quickly calculate the products $bmod 12$ across this set:
beginarraycc
p backslash q & 1 & 5 & 7 & 11 \ hline
1 & 1 & 5 & 7 & 11 \
5 & 5 & 1 & 11 & 7 \
7 & 7 & 11 & 1 & 5 \
11 & 11 & 7 & 5 & 1 \
endarray
We are looking for $p,q$ pairs that give $11equiv -1 bmod 12$ in the above table so that $12mid pq+1$. So we would select the pairs from the different residue class sets appropriately, which gives a way to calculate how many options in total.
For example, the set of primes in range $equiv 1 bmod 12$ (call this $S_1$) consists of just $13,37$. This gives us $|S_1|=2$ and similarly $|S_5|=4,$ $|S_7|=4,$ and $|S_11|=3$. Thus we will have $2cdot 3 + 4cdot 4 + 4cdot 4 + 3cdot 2 = 2cdot 6+2cdot 16= fbox44 $ ordered pair solutions.
answered Jul 24 at 16:49
Joffan
31.8k43169
But I guess 1 is not a prime number.
– mampuuu
Jul 24 at 18:48
No it isn't; you're looking (in that case) for the prime numbers that fit $pequiv 1 bmod 12$. Similarly in the other cases you're looking for the sets of prime numbers $equiv 5,7,11 bmod 12$. Example added to answer.
– Joffan
Jul 24 at 18:51
add a comment |Â
up vote
0
down vote
Clearly we need both $p$ and $q$ to be coprime to $12$, so we cannot have $p$ or $q$ equal to either $2$ or $3$.
This means that $p,qin1,5,7,11bmod 12$. We can quickly calculate the products $bmod 12$ across this set:
beginarraycc
p backslash q & 1 & 5 & 7 & 11 \ hline
1 & 1 & 5 & 7 & 11 \
5 & 5 & 1 & 11 & 7 \
7 & 7 & 11 & 1 & 5 \
11 & 11 & 7 & 5 & 1 \
endarray
We are looking for $p,q$ pairs that give $11equiv -1 bmod 12$ in the above table so that $12mid pq+1$. So we would select the pairs from the different residue class sets appropriately, which gives a way to calculate how many options in total.
For example, the set of primes in range $equiv 1 bmod 12$ (call this $S_1$) consists of just $13,37$. This gives us $|S_1|=2$ and similarly $|S_5|=4,$ $|S_7|=4,$ and $|S_11|=3$. Thus we will have $2cdot 3 + 4cdot 4 + 4cdot 4 + 3cdot 2 = 2cdot 6+2cdot 16= fbox44 $ ordered pair solutions.
answered Jul 24 at 16:49
Joffan
31.8k43169
But I guess 1 is not a prime number.
– mampuuu
Jul 24 at 18:48
No it isn't; you're looking (in that case) for the prime numbers that fit $pequiv 1 bmod 12$. Similarly in the other cases you're looking for the sets of prime numbers $equiv 5,7,11 bmod 12$. Example added to answer.
– Joffan
Jul 24 at 18:51
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Clearly we need both $p$ and $q$ to be coprime to $12$, so we cannot have $p$ or $q$ equal to either $2$ or $3$.
This means that $p,qin1,5,7,11bmod 12$. We can quickly calculate the products $bmod 12$ across this set:
beginarraycc
p backslash q & 1 & 5 & 7 & 11 \ hline
1 & 1 & 5 & 7 & 11 \
5 & 5 & 1 & 11 & 7 \
7 & 7 & 11 & 1 & 5 \
11 & 11 & 7 & 5 & 1 \
endarray
We are looking for $p,q$ pairs that give $11equiv -1 bmod 12$ in the above table so that $12mid pq+1$. So we would select the pairs from the different residue class sets appropriately, which gives a way to calculate how many options in total.
For example, the set of primes in range $equiv 1 bmod 12$ (call this $S_1$) consists of just $13,37$. This gives us $|S_1|=2$ and similarly $|S_5|=4,$ $|S_7|=4,$ and $|S_11|=3$. Thus we will have $2cdot 3 + 4cdot 4 + 4cdot 4 + 3cdot 2 = 2cdot 6+2cdot 16= fbox44 $ ordered pair solutions.
answered Jul 24 at 16:49
Joffan
31.8k43169
Clearly we need both $p$ and $q$ to be coprime to $12$, so we cannot have $p$ or $q$ equal to either $2$ or $3$.
This means that $p,qin1,5,7,11bmod 12$. We can quickly calculate the products $bmod 12$ across this set:
beginarraycc
p backslash q & 1 & 5 & 7 & 11 \ hline
1 & 1 & 5 & 7 & 11 \
5 & 5 & 1 & 11 & 7 \
7 & 7 & 11 & 1 & 5 \
11 & 11 & 7 & 5 & 1 \
endarray
We are looking for $p,q$ pairs that give $11equiv -1 bmod 12$ in the above table so that $12mid pq+1$. So we would select the pairs from the different residue class sets appropriately, which gives a way to calculate how many options in total.
For example, the set of primes in range $equiv 1 bmod 12$ (call this $S_1$) consists of just $13,37$. This gives us $|S_1|=2$ and similarly $|S_5|=4,$ $|S_7|=4,$ and $|S_11|=3$. Thus we will have $2cdot 3 + 4cdot 4 + 4cdot 4 + 3cdot 2 = 2cdot 6+2cdot 16= fbox44 $ ordered pair solutions.
answered Jul 24 at 16:49
Joffan
31.8k43169
edited Jul 26 at 16:13
answered Jul 24 at 16:49
Joffan
31.8k43169
answered Jul 24 at 16:49
Joffan
31.8k43169
answered Jul 24 at 16:49
Joffan
31.8k43169
31.8k43169
But I guess 1 is not a prime number.
– mampuuu
Jul 24 at 18:48
No it isn't; you're looking (in that case) for the prime numbers that fit $pequiv 1 bmod 12$. Similarly in the other cases you're looking for the sets of prime numbers $equiv 5,7,11 bmod 12$. Example added to answer.
– Joffan
Jul 24 at 18:51
add a comment |Â
But I guess 1 is not a prime number.
– mampuuu
Jul 24 at 18:48
No it isn't; you're looking (in that case) for the prime numbers that fit $pequiv 1 bmod 12$. Similarly in the other cases you're looking for the sets of prime numbers $equiv 5,7,11 bmod 12$. Example added to answer.
– Joffan
Jul 24 at 18:51
But I guess 1 is not a prime number.
– mampuuu
Jul 24 at 18:48
But I guess 1 is not a prime number.
– mampuuu
Jul 24 at 18:48
No it isn't; you're looking (in that case) for the prime numbers that fit $pequiv 1 bmod 12$. Similarly in the other cases you're looking for the sets of prime numbers $equiv 5,7,11 bmod 12$. Example added to answer.
– Joffan
Jul 24 at 18:51
No it isn't; you're looking (in that case) for the prime numbers that fit $pequiv 1 bmod 12$. Similarly in the other cases you're looking for the sets of prime numbers $equiv 5,7,11 bmod 12$. Example added to answer.
– Joffan
Jul 24 at 18:51
add a comment |Â
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1
What have you tried in solving the question? Please include that below your question body (as an edit).
– Parcly Taxel
Jul 24 at 15:53
you could do this in excel by hand -- what is the problem exactly?
– gt6989b
Jul 24 at 15:56
@gt6989b Or just by hand, without excel ...
– Peter
Jul 24 at 15:57
@Peter how? There are going to be a lot of combinations
– mampuuu
Jul 24 at 15:59
We have $15$ primes below $50$, making $105$ pairs. Moreover $2$ and $3$ can be omitted because a number of the form $2k+1$ is odd and a number of the form $3k+1$ not disivisble by $3$. So, the number of pairs decreases to $78$. With a bit patience, well feasible by hand.
– Peter
Jul 24 at 16:03