Writing integers as the product of prime powers
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 Nov 19 '18 at 6:25
Brahadeesh
6,11742360
asked Nov 19 '18 at 6:03
Raul Quintanilla Jr.
702
add a comment |
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 Nov 19 '18 at 6:25
Brahadeesh
6,11742360
asked Nov 19 '18 at 6:03
Raul Quintanilla Jr.
702
1
Take $p=7, q=11$ for example. Can you write $n=2$ as $p^{k}q^{j}$?.
– Kavi Rama Murthy
Nov 19 '18 at 6:07
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
Nov 19 '18 at 7:20
add a comment |
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 Nov 19 '18 at 6:25
Brahadeesh
6,11742360
asked Nov 19 '18 at 6:03
Raul Quintanilla Jr.
702
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 Nov 19 '18 at 6:25
Brahadeesh
6,11742360
asked Nov 19 '18 at 6:03
Raul Quintanilla Jr.
702
edited Nov 19 '18 at 6:25
Brahadeesh
6,11742360
asked Nov 19 '18 at 6:03
Raul Quintanilla Jr.
702
edited Nov 19 '18 at 6:25
Brahadeesh
6,11742360
edited Nov 19 '18 at 6:25
Brahadeesh
6,11742360
edited Nov 19 '18 at 6:25
Brahadeesh
6,11742360
6,11742360
asked Nov 19 '18 at 6:03
Raul Quintanilla Jr.
702
asked Nov 19 '18 at 6:03
Raul Quintanilla Jr.
702
asked Nov 19 '18 at 6:03
Raul Quintanilla Jr.
702
702
1
Take $p=7, q=11$ for example. Can you write $n=2$ as $p^{k}q^{j}$?.
– Kavi Rama Murthy
Nov 19 '18 at 6:07
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
Nov 19 '18 at 7:20
add a comment |
1
Take $p=7, q=11$ for example. Can you write $n=2$ as $p^{k}q^{j}$?.
– Kavi Rama Murthy
Nov 19 '18 at 6:07
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
Nov 19 '18 at 7:20
1
1
Take $p=7, q=11$ for example. Can you write $n=2$ as $p^{k}q^{j}$?.
– Kavi Rama Murthy
Nov 19 '18 at 6:07
Take $p=7, q=11$ for example. Can you write $n=2$ as $p^{k}q^{j}$?.
– Kavi Rama Murthy
Nov 19 '18 at 6:07
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
Nov 19 '18 at 7:20
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
Nov 19 '18 at 7:20
add a comment |
2 Answers
2
active
oldest
votes
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 Nov 19 '18 at 7:10
fleablood
68.2k22684
fleablood.Very nice!
– Peter Szilas
Nov 19 '18 at 7:17
add a comment |
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 Nov 19 '18 at 6:24
Ross Millikan
291k23196371
2
Actually $16$ fails as $16=2^4$ and $4>3$. But it's the only one that fails.
– fleablood
Nov 19 '18 at 7:11
@fleablood: I missed the restriction on the exponents, just thought we could only have two prime factors. I'll fix.
– Ross Millikan
Nov 19 '18 at 14:43
add a comment |
Your Answer
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2 Answers
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2 Answers
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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 Nov 19 '18 at 7:10
fleablood
68.2k22684
fleablood.Very nice!
– Peter Szilas
Nov 19 '18 at 7:17
add a comment |
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 Nov 19 '18 at 7:10
fleablood
68.2k22684
fleablood.Very nice!
– Peter Szilas
Nov 19 '18 at 7:17
add a comment |
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 Nov 19 '18 at 7:10
fleablood
68.2k22684
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 Nov 19 '18 at 7:10
fleablood
68.2k22684
edited Nov 19 '18 at 7:22
answered Nov 19 '18 at 7:10
fleablood
68.2k22684
answered Nov 19 '18 at 7:10
fleablood
68.2k22684
answered Nov 19 '18 at 7:10
fleablood
68.2k22684
68.2k22684
fleablood.Very nice!
– Peter Szilas
Nov 19 '18 at 7:17
add a comment |
fleablood.Very nice!
– Peter Szilas
Nov 19 '18 at 7:17
fleablood.Very nice!
– Peter Szilas
Nov 19 '18 at 7:17
fleablood.Very nice!
– Peter Szilas
Nov 19 '18 at 7:17
add a comment |
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 Nov 19 '18 at 6:24
Ross Millikan
291k23196371
2
Actually $16$ fails as $16=2^4$ and $4>3$. But it's the only one that fails.
– fleablood
Nov 19 '18 at 7:11
@fleablood: I missed the restriction on the exponents, just thought we could only have two prime factors. I'll fix.
– Ross Millikan
Nov 19 '18 at 14:43
add a comment |
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 Nov 19 '18 at 6:24
Ross Millikan
291k23196371
2
Actually $16$ fails as $16=2^4$ and $4>3$. But it's the only one that fails.
– fleablood
Nov 19 '18 at 7:11
@fleablood: I missed the restriction on the exponents, just thought we could only have two prime factors. I'll fix.
– Ross Millikan
Nov 19 '18 at 14:43
add a comment |
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 Nov 19 '18 at 6:24
Ross Millikan
291k23196371
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 Nov 19 '18 at 6:24
Ross Millikan
291k23196371
edited Nov 19 '18 at 14:44
answered Nov 19 '18 at 6:24
Ross Millikan
291k23196371
answered Nov 19 '18 at 6:24
Ross Millikan
291k23196371
answered Nov 19 '18 at 6:24
Ross Millikan
291k23196371
291k23196371
2
Actually $16$ fails as $16=2^4$ and $4>3$. But it's the only one that fails.
– fleablood
Nov 19 '18 at 7:11
@fleablood: I missed the restriction on the exponents, just thought we could only have two prime factors. I'll fix.
– Ross Millikan
Nov 19 '18 at 14:43
add a comment |
2
Actually $16$ fails as $16=2^4$ and $4>3$. But it's the only one that fails.
– fleablood
Nov 19 '18 at 7:11
@fleablood: I missed the restriction on the exponents, just thought we could only have two prime factors. I'll fix.
– Ross Millikan
Nov 19 '18 at 14:43
2
2
Actually $16$ fails as $16=2^4$ and $4>3$. But it's the only one that fails.
– fleablood
Nov 19 '18 at 7:11
Actually $16$ fails as $16=2^4$ and $4>3$. But it's the only one that fails.
– fleablood
Nov 19 '18 at 7:11
@fleablood: I missed the restriction on the exponents, just thought we could only have two prime factors. I'll fix.
– Ross Millikan
Nov 19 '18 at 14:43
@fleablood: I missed the restriction on the exponents, just thought we could only have two prime factors. I'll fix.
– Ross Millikan
Nov 19 '18 at 14:43
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
Nov 19 '18 at 6:07
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
Nov 19 '18 at 7:20