Question

Prove the divisibility of the following numbers:

Answer 1

Answer:

$5^{12}$

Step-by-step explanation:

$25^{7} +5^{13}$ by 30 is NOT no solution

Answer 2

Make use of prime factorizations:

$16^5+2^{15}=(2^4)^5+2^{15}=2^{20}+2^{15}$

Both terms have a common factor of $2^{15}$:

$16^5+2^{15}=2^{15}\left(2^5+1\right)=2^{15}\cdot33$

- - -

The second one is not true! We can write

$15^7+5^{13}=(3\cdot5)^7+5^{13}=3^7\cdot5^7+5^{13}$

Both terms have a common factor of $5^7$:

$15^7+5^{13}=5^7\left(3^7+5^6\right)$

Since $30=5\cdot6$, and $5\mid5^7$, we'd still have to show that $5^6(3^7+5^6)$ is a multiple of 6. This is impossible, because $6=3\cdot2$ and there is no multiple of 2 that can be factored out.