If sum is $\frac{p(p+1)}{2}=2^m3^n (m+n)$ then $1000 \leq 2^m 3^n (m+n) \leq 3^{m+n} (m+n) \to m+n \geq 5$ $10000 > 2^m 3^n (m+n)\geq 2^{m+n} (m+n) \to m+n \leq 9$ And as $2,3 \not |m+n \to m+n=5,7$ $m+n=5$ Then only possible variant is $3^5*5$ but $\frac{p(p+1)}{2}=3^5*5$ has no solutions $m+n=7$ $p(p+1)=7*2^{m+1} 3^n$ and $2000 \leq p(p+1) <20000 \to 45 \leq p \leq 140$ One of $p$ or $p+1$ is in form $7*2^a*3^b$ so possible variants are $56,63,84,112,126$ but as other number from pair $\{p,p+1\}$ should be in form $2^c 3^d$ then only $p=63$ is suitable And $p(p+1)=63*64=2*2^5*3^2*7$ is solution