Question

Prove the divisibility of the following numbers:

PLEASE HELP! I HAVE NO IDEA... ANY HELP IS APPRECIATED!!!!!!

1. 16^5 + 2^15 by 33

2. 15^7 + 5^13 by 30

Answer 1

`5^(12)`

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.