Mixing
Writing integers as the product of prime powers
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Let $p,q$ be primes such that $p lt q$.
Show that every positive integer $n lt 15$ can be written in the form $p^kq^j$ for $k,j = {0,1,2,3}$.
I know that every integer can be written as the product of primes (like $12 = 2^2 * 3^1$) but I'm not sure how to make a coherent proof for it.
I tried separating all values for $n$ into even, odd, and prime cases but I couldn't figure out where to take that. Any help would be appreciated.
elementary-number-theory prime-numbers
edited yesterday
Brahadeesh
5,53941956
asked yesterday
Raul Quintanilla Jr.
442
add a comment |
up vote
0
down vote
favorite
Let $p,q$ be primes such that $p lt q$.
Show that every positive integer $n lt 15$ can be written in the form $p^kq^j$ for $k,j = {0,1,2,3}$.
I know that every integer can be written as the product of primes (like $12 = 2^2 * 3^1$) but I'm not sure how to make a coherent proof for it.
I tried separating all values for $n$ into even, odd, and prime cases but I couldn't figure out where to take that. Any help would be appreciated.
elementary-number-theory prime-numbers
edited yesterday
Brahadeesh
5,53941956
asked yesterday
Raul Quintanilla Jr.
442
1
Take $p=7, q=11$ for example. Can you write $n=2$ as $p^{k}q^{j}$?.
– Kavi Rama Murthy
yesterday
Are you sure you are writing the question correctly? The fact that $p <q $ seems irrelevant. If $p$ and $q $ need to be the same for all numbers it's not true ($14$ needs 2 and $7$ and $6$ needs $2$ and $3$) and if $p,q $ can change its easy to just list the prime factors of all (and set a second prime to $0$ power if there is only one prime factor.
– fleablood
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $p,q$ be primes such that $p lt q$.
Show that every positive integer $n lt 15$ can be written in the form $p^kq^j$ for $k,j = {0,1,2,3}$.
I know that every integer can be written as the product of primes (like $12 = 2^2 * 3^1$) but I'm not sure how to make a coherent proof for it.
I tried separating all values for $n$ into even, odd, and prime cases but I couldn't figure out where to take that. Any help would be appreciated.
elementary-number-theory prime-numbers
edited yesterday
Brahadeesh
5,53941956
asked yesterday
Raul Quintanilla Jr.
442
Let $p,q$ be primes such that $p lt q$.
Show that every positive integer $n lt 15$ can be written in the form $p^kq^j$ for $k,j = {0,1,2,3}$.
I know that every integer can be written as the product of primes (like $12 = 2^2 * 3^1$) but I'm not sure how to make a coherent proof for it.
I tried separating all values for $n$ into even, odd, and prime cases but I couldn't figure out where to take that. Any help would be appreciated.
elementary-number-theory prime-numbers
elementary-number-theory prime-numbers
edited yesterday
Brahadeesh
5,53941956
asked yesterday
Raul Quintanilla Jr.
442
edited yesterday
Brahadeesh
5,53941956
asked yesterday
Raul Quintanilla Jr.
442
edited yesterday
Brahadeesh
5,53941956
edited yesterday
Brahadeesh
5,53941956
edited yesterday
Brahadeesh
5,53941956
5,53941956
asked yesterday
Raul Quintanilla Jr.
442
asked yesterday
Raul Quintanilla Jr.
442
asked yesterday
Raul Quintanilla Jr.
442
442
1
Take $p=7, q=11$ for example. Can you write $n=2$ as $p^{k}q^{j}$?.
– Kavi Rama Murthy
yesterday
Are you sure you are writing the question correctly? The fact that $p <q $ seems irrelevant. If $p$ and $q $ need to be the same for all numbers it's not true ($14$ needs 2 and $7$ and $6$ needs $2$ and $3$) and if $p,q $ can change its easy to just list the prime factors of all (and set a second prime to $0$ power if there is only one prime factor.
– fleablood
yesterday
add a comment |
1
Take $p=7, q=11$ for example. Can you write $n=2$ as $p^{k}q^{j}$?.
– Kavi Rama Murthy
yesterday
Are you sure you are writing the question correctly? The fact that $p <q $ seems irrelevant. If $p$ and $q $ need to be the same for all numbers it's not true ($14$ needs 2 and $7$ and $6$ needs $2$ and $3$) and if $p,q $ can change its easy to just list the prime factors of all (and set a second prime to $0$ power if there is only one prime factor.
– fleablood
yesterday
1
1
Take $p=7, q=11$ for example. Can you write $n=2$ as $p^{k}q^{j}$?.
– Kavi Rama Murthy
yesterday
Take $p=7, q=11$ for example. Can you write $n=2$ as $p^{k}q^{j}$?.
– Kavi Rama Murthy
yesterday
Are you sure you are writing the question correctly? The fact that $p <q $ seems irrelevant. If $p$ and $q $ need to be the same for all numbers it's not true ($14$ needs 2 and $7$ and $6$ needs $2$ and $3$) and if $p,q $ can change its easy to just list the prime factors of all (and set a second prime to $0$ power if there is only one prime factor.
– fleablood
yesterday
Are you sure you are writing the question correctly? The fact that $p <q $ seems irrelevant. If $p$ and $q $ need to be the same for all numbers it's not true ($14$ needs 2 and $7$ and $6$ needs $2$ and $3$) and if $p,q $ can change its easy to just list the prime factors of all (and set a second prime to $0$ power if there is only one prime factor.
– fleablood
yesterday
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
As $j,k $ can be zero, $1$ and numbers with only one prime factor, can be written as $p^0q^0$ and $p^kq^0$.
For a number not to be able to be written this was either the number must have at least three prime factors which would mean it is at least $2*3*5=30$, or it will have a prime power of at least $4$ which would mean it is at least $2^4=16$.
Thus all $nle 15$ can be so written.
Anyway, even if you couldn't come up with the idea of that proof there is no reason you wouldn't just DO them.
There are only 14 to do and they are ALL exceedingly EASY.
$$1=2^03^0;2=2^13^0;3=2^03^1;4=2^23^0;5=2^05^1;6=2^13^1.... $$ do I need to go on?
Addendum. As $3^4=81>30$ and $2^5=32>30$ the only number less than $30$ that can't be so written is $16$. Between $17$ and $60$ the only numbers that can't are $32=2^5$ and $48=3*2^4$.
answered yesterday
fleablood
65.4k22682
fleablood.Very nice!
– Peter Szilas
yesterday
add a comment |
up vote
0
down vote
There are only $14$ positive integers less than $15$. Just show a factorization of the requested kind for each one and claim victory. The statement first fails for $16=2^4$ because the exponent is too high, then for $30=2cdot 3 cdot 5$ because there are three different prime factors.
answered yesterday
Ross Millikan
287k23195364
2
Actually $16$ fails as $16=2^4$ and $4>3$. But it's the only one that fails.
– fleablood
yesterday
@fleablood: I missed the restriction on the exponents, just thought we could only have two prime factors. I'll fix.
– Ross Millikan
19 hours ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
As $j,k $ can be zero, $1$ and numbers with only one prime factor, can be written as $p^0q^0$ and $p^kq^0$.
For a number not to be able to be written this was either the number must have at least three prime factors which would mean it is at least $2*3*5=30$, or it will have a prime power of at least $4$ which would mean it is at least $2^4=16$.
Thus all $nle 15$ can be so written.
Anyway, even if you couldn't come up with the idea of that proof there is no reason you wouldn't just DO them.
There are only 14 to do and they are ALL exceedingly EASY.
$$1=2^03^0;2=2^13^0;3=2^03^1;4=2^23^0;5=2^05^1;6=2^13^1.... $$ do I need to go on?
Addendum. As $3^4=81>30$ and $2^5=32>30$ the only number less than $30$ that can't be so written is $16$. Between $17$ and $60$ the only numbers that can't are $32=2^5$ and $48=3*2^4$.
answered yesterday
fleablood
65.4k22682
fleablood.Very nice!
– Peter Szilas
yesterday
add a comment |
up vote
1
down vote
As $j,k $ can be zero, $1$ and numbers with only one prime factor, can be written as $p^0q^0$ and $p^kq^0$.
For a number not to be able to be written this was either the number must have at least three prime factors which would mean it is at least $2*3*5=30$, or it will have a prime power of at least $4$ which would mean it is at least $2^4=16$.
Thus all $nle 15$ can be so written.
Anyway, even if you couldn't come up with the idea of that proof there is no reason you wouldn't just DO them.
There are only 14 to do and they are ALL exceedingly EASY.
$$1=2^03^0;2=2^13^0;3=2^03^1;4=2^23^0;5=2^05^1;6=2^13^1.... $$ do I need to go on?
Addendum. As $3^4=81>30$ and $2^5=32>30$ the only number less than $30$ that can't be so written is $16$. Between $17$ and $60$ the only numbers that can't are $32=2^5$ and $48=3*2^4$.
answered yesterday
fleablood
65.4k22682
fleablood.Very nice!
– Peter Szilas
yesterday
add a comment |
up vote
1
down vote
up vote
1
down vote
As $j,k $ can be zero, $1$ and numbers with only one prime factor, can be written as $p^0q^0$ and $p^kq^0$.
For a number not to be able to be written this was either the number must have at least three prime factors which would mean it is at least $2*3*5=30$, or it will have a prime power of at least $4$ which would mean it is at least $2^4=16$.
Thus all $nle 15$ can be so written.
Anyway, even if you couldn't come up with the idea of that proof there is no reason you wouldn't just DO them.
There are only 14 to do and they are ALL exceedingly EASY.
$$1=2^03^0;2=2^13^0;3=2^03^1;4=2^23^0;5=2^05^1;6=2^13^1.... $$ do I need to go on?
Addendum. As $3^4=81>30$ and $2^5=32>30$ the only number less than $30$ that can't be so written is $16$. Between $17$ and $60$ the only numbers that can't are $32=2^5$ and $48=3*2^4$.
answered yesterday
fleablood
65.4k22682
As $j,k $ can be zero, $1$ and numbers with only one prime factor, can be written as $p^0q^0$ and $p^kq^0$.
For a number not to be able to be written this was either the number must have at least three prime factors which would mean it is at least $2*3*5=30$, or it will have a prime power of at least $4$ which would mean it is at least $2^4=16$.
Thus all $nle 15$ can be so written.
Anyway, even if you couldn't come up with the idea of that proof there is no reason you wouldn't just DO them.
There are only 14 to do and they are ALL exceedingly EASY.
$$1=2^03^0;2=2^13^0;3=2^03^1;4=2^23^0;5=2^05^1;6=2^13^1.... $$ do I need to go on?
Addendum. As $3^4=81>30$ and $2^5=32>30$ the only number less than $30$ that can't be so written is $16$. Between $17$ and $60$ the only numbers that can't are $32=2^5$ and $48=3*2^4$.
answered yesterday
fleablood
65.4k22682
edited yesterday
answered yesterday
fleablood
65.4k22682
answered yesterday
fleablood
65.4k22682
answered yesterday
fleablood
65.4k22682
65.4k22682
fleablood.Very nice!
– Peter Szilas
yesterday
add a comment |
fleablood.Very nice!
– Peter Szilas
yesterday
fleablood.Very nice!
– Peter Szilas
yesterday
fleablood.Very nice!
– Peter Szilas
yesterday
add a comment |
up vote
0
down vote
There are only $14$ positive integers less than $15$. Just show a factorization of the requested kind for each one and claim victory. The statement first fails for $16=2^4$ because the exponent is too high, then for $30=2cdot 3 cdot 5$ because there are three different prime factors.
answered yesterday
Ross Millikan
287k23195364
2
Actually $16$ fails as $16=2^4$ and $4>3$. But it's the only one that fails.
– fleablood
yesterday
@fleablood: I missed the restriction on the exponents, just thought we could only have two prime factors. I'll fix.
– Ross Millikan
19 hours ago
add a comment |
up vote
0
down vote
There are only $14$ positive integers less than $15$. Just show a factorization of the requested kind for each one and claim victory. The statement first fails for $16=2^4$ because the exponent is too high, then for $30=2cdot 3 cdot 5$ because there are three different prime factors.
answered yesterday
Ross Millikan
287k23195364
2
Actually $16$ fails as $16=2^4$ and $4>3$. But it's the only one that fails.
– fleablood
yesterday
@fleablood: I missed the restriction on the exponents, just thought we could only have two prime factors. I'll fix.
– Ross Millikan
19 hours ago
add a comment |
up vote
0
down vote
up vote
0
down vote
There are only $14$ positive integers less than $15$. Just show a factorization of the requested kind for each one and claim victory. The statement first fails for $16=2^4$ because the exponent is too high, then for $30=2cdot 3 cdot 5$ because there are three different prime factors.
answered yesterday
Ross Millikan
287k23195364
There are only $14$ positive integers less than $15$. Just show a factorization of the requested kind for each one and claim victory. The statement first fails for $16=2^4$ because the exponent is too high, then for $30=2cdot 3 cdot 5$ because there are three different prime factors.
answered yesterday
Ross Millikan
287k23195364
edited 19 hours ago
answered yesterday
Ross Millikan
287k23195364
answered yesterday
Ross Millikan
287k23195364
answered yesterday
Ross Millikan
287k23195364
287k23195364
2
Actually $16$ fails as $16=2^4$ and $4>3$. But it's the only one that fails.
– fleablood
yesterday
@fleablood: I missed the restriction on the exponents, just thought we could only have two prime factors. I'll fix.
– Ross Millikan
19 hours ago
add a comment |
2
Actually $16$ fails as $16=2^4$ and $4>3$. But it's the only one that fails.
– fleablood
yesterday
@fleablood: I missed the restriction on the exponents, just thought we could only have two prime factors. I'll fix.
– Ross Millikan
19 hours ago
2
2
Actually $16$ fails as $16=2^4$ and $4>3$. But it's the only one that fails.
– fleablood
yesterday
Actually $16$ fails as $16=2^4$ and $4>3$. But it's the only one that fails.
– fleablood
yesterday
@fleablood: I missed the restriction on the exponents, just thought we could only have two prime factors. I'll fix.
– Ross Millikan
19 hours ago
@fleablood: I missed the restriction on the exponents, just thought we could only have two prime factors. I'll fix.
– Ross Millikan
19 hours ago
add a comment |
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1
Take $p=7, q=11$ for example. Can you write $n=2$ as $p^{k}q^{j}$?.
– Kavi Rama Murthy
yesterday
Are you sure you are writing the question correctly? The fact that $p <q $ seems irrelevant. If $p$ and $q $ need to be the same for all numbers it's not true ($14$ needs 2 and $7$ and $6$ needs $2$ and $3$) and if $p,q $ can change its easy to just list the prime factors of all (and set a second prime to $0$ power if there is only one prime factor.
– fleablood
yesterday