Writing integers as the product of prime powers












0














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







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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
















0














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







share|cite|improve this question




















  • 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














0












0








0







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 19 '18 at 6:25









Brahadeesh

6,11742360




6,11742360










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














  • 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










2 Answers
2






active

oldest

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1














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 $48=3*2^4$.






share|cite|improve this answer























  • fleablood.Very nice!
    – Peter Szilas
    Nov 19 '18 at 7:17



















0














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.






share|cite|improve this answer



















  • 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











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2 Answers
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2 Answers
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1














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 $48=3*2^4$.






share|cite|improve this answer























  • fleablood.Very nice!
    – Peter Szilas
    Nov 19 '18 at 7:17
















1














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 $48=3*2^4$.






share|cite|improve this answer























  • fleablood.Very nice!
    – Peter Szilas
    Nov 19 '18 at 7:17














1












1








1






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 $48=3*2^4$.






share|cite|improve this answer














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 $48=3*2^4$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 19 '18 at 7:22

























answered Nov 19 '18 at 7:10









fleablood

68.2k22684




68.2k22684












  • 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




fleablood.Very nice!
– Peter Szilas
Nov 19 '18 at 7:17











0














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.






share|cite|improve this answer



















  • 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
















0














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.






share|cite|improve this answer



















  • 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














0












0








0






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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 19 '18 at 14:44

























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














  • 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


















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