Statement 1:

Factors into k*m*(m^2-1)=k*m(m-1)(m+1). The factors will always be (m-1), m, and (m+1), meaning one of them will be divisible by 3. I don't see any way to determine if it's divisible by 10.

I can show that Statement 2 will always be divisible by 10:

We want the last digit to be divisible by 10. The last digit of n^5 will always be its first digit, then when you subtract n from that, the result's last digit will always be 0, thus it will be divisible by 10:

last digit of n=1: 1, 1, 1, 1, 1

last digit of n=2: 2, 4, 8, 6, 2

last digit of n=3: 3, 9, 7, 1, 3

last digit of n=4: 4, 6, 4, 6, 4

last digit of n=5: 5, 5, 5, 5, 5

last digit of n=6: 6, 6, 6, 6, 6

last digit of n=7: 7, 9, 3, 1, 7

last digit of n=8: 8, 4, 2, 6, 8

last digit of n=9: 9, 1, 9, 1, 9

We can also factor the right side of the equation: n(n^4-1), which can be further broken into: n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2+1). Once again, it's divisible by 3 because n, (n-1), and (n+1) are all factors, one of which must be divisible by 3. Maybe this can somehow be used to prove divisibility by 10, but I don't see it.

Thus, 2 will be sufficient.

B

PS, are Manhattan problems supposed to be doable in 2 minutes? I doubt if there are more than 10 people in the world who can solve this one in 2 minutes.