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.
On
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 $n\le 15$ 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 $48=3*2^4$.
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 $30=2\cdot 3 \cdot 5$ because there are three different prime factors.