You can
put this solution on YOUR website!I'm not at all sure how to even approach the first one. Perhaps someone else can help you.
But here is a rather intuitive discussion of the second problem:
If some integer p is divided by another integer m,

, using integer division, then there are exactly

possible non-zero remainders. If given m consectutive integers, dividing each by m will result in m different remainders. Since there are only

non-zero remainders available, one of the m remainders must be 0, hence one of the m integers is evenly divisible by m.