This is bound to be a somewhat controversial question because the distribution of the item prices is not specified, and in the "real world" the number of cents is not uniformly distributed. For example, a price like xx.98 or xx.99 is much more likely than a price like xx.00, or at least so it seems to me.

Nonetheless, I think by any "reasonable definition" it will turn out that the number of cents in the total is approximately distributed as a uniform discrete distribution having parameters a=0, b=99, n=100 (see Uniform distribution (discrete - Wikipedia, the free encyclopedia)), so the probability that the number of cents will be zero is approximately 1/100.

If you insist, a more detailed approach is to assume each individual item has a price in which the number of cents has a uniform discrete distribution with a=0, b=99, n=100, which has a mean of 49.5 and a variance of 833.25. The sum of the cents therefore has an approximate normal distribution with mean 13 * 49.5 and a variance of 13 * 833.25. The sum of the cents is between 0 and 1287 (inclusive), so we can use the normal approximation to compute the probability that the sum is 0, 100, 200, 300, ..., 1200. If we add these probabilities up we then find that the total is 0.01, as predicted.