First, let's find the probability mass function (PMF) of Y.
Since X is uniformly distributed on [0, k], the probability that X takes on a value in the interval [n, n+1) is equal to the length of that interval divided by the length of the entire interval [0, k].
The length of the interval [n, n+1) is 1, and the length of [0, k] is k, so the probability that X takes on a value in [n, n+1) is 1/k.
Therefore, p_Y(y), the probability that Y is equal to y, is equal to the probability that X takes on a value in the interval [y, y+1), which is 1/k.
Since p_Y(y) = 0 for y ≥ ℓ + 1, we want to find the largest integer ℓ such that p_Y(y) ≠ 0 for y = 0, 1, ..., ℓ.
For y = 0, 1, ..., ℓ, p_Y(y) = 1/k ≠ 0.
Therefore, ℓ = k - 1.
So we have ℓ = k - 1 and p_Y(y) = 1/k for y = 0, 1, ..., k - 1.
Next, let's find the probability density function (PDF) of Z.
Since Z = frac(X), Z has a uniform distribution on [0, 1).
The PDF of a uniform distribution on [0, 1) is constant over its support (in this case, [0, 1)) and is equal to 1/c, where c is the length of the support.
Since the support of Z is [0, 1), c = 1.
Therefore, c = 1 and f_Z(z) = 1 for z ∈ (0, 1).
So we have c = 1 and f_Z(z) = 1 for z ∈ (0, 1).
In summary, the values we found are:
ℓ = k - 1
p_Y(y) = 1/k for y = 0, 1, ..., k - 1
c = 1
f_Z(z) = 1 for z ∈ (0, 1)