Make use of prime factorizations:
[tex]16^5+2^{15}=(2^4)^5+2^{15}=2^{20}+2^{15}[/tex]
Both terms have a common factor of [tex]2^{15}[/tex]:
[tex]16^5+2^{15}=2^{15}\left(2^5+1\right)=2^{15}\cdot33[/tex]
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The second one is not true! We can write
[tex]15^7+5^{13}=(3\cdot5)^7+5^{13}=3^7\cdot5^7+5^{13}[/tex]
Both terms have a common factor of [tex]5^7[/tex]:
[tex]15^7+5^{13}=5^7\left(3^7+5^6\right)[/tex]
Since [tex]30=5\cdot6[/tex], and [tex]5\mid5^7[/tex], we'd still have to show that [tex]5^6(3^7+5^6)[/tex] is a multiple of 6. This is impossible, because [tex]6=3\cdot2[/tex] and there is no multiple of 2 that can be factored out.
