2 short math proofs - numbers in base 5

[Pompey]

I'm having real trouble getting started - any help or guidance will be much appreciated.

1) Prove that a number in base 5 is even iff the sum of its digits is even.

2) Prove that 5^k = 3q +1 (for k even) = 3q -1 (for k odd) (k and q are natural numbers)

Google's not been helpful in finding a method to tackle these.

[bigpooch]

In both questions, simply use modular arithmetic. Let
sigma (k=1 to n){E(k) } denote summation with index k over
the expression E(k).

Recall or note that if x, y are integers and m some modulus,
if ~ denotes congruence (mod m), then

x~a and y~b implies xy~ab.

Thus, also x^k ~ a^k for any positive integer k.

1) Use m=2 and let ~ denote congruence (mod 2). Suppose
a number N = sigma(k=0 to n) {d(k)[b^k]} (base b
representation with "base b digits" d(k) ). If b is any
odd base, such as 5, then b~1 (mod 2), so that
N ~ sigma(k=0 to n) {d(k)}{1^k}
~sigma(k=0 to n) {d(k)}. So, being odd/even depends on the
sum of the digits. [Note b^0=1 ~ 1.]

Thus, a number in base b, where b>=3 is an odd integer, is
even iff the sum of digits is even.

One can also see that for any even base, a number written
in that base is even iff the last "digit" is even.

2) Use m=3 and let ~ denote congruence (mod 3). Then 5 ~ -1
and 5^k ~ (-1)^k and the result is clear.

[Pompey]

Thanks bigpooch, it all fell into place when I saw 5 =(Congruent to)= -1 (mod 3)