Good afternoon Candice! Here are my workings for part ii. We note that for every one “2” paired with one “5”, we obtain one “10”. Note that the number of “10”s which we can obtain is limited to the available number of “2”s or “5”s. So for example, 2^4 * 5 = 80 will only contain one “10” because there is only one “5” (and therefore no more “10”s can be generated).

As such, I found that there are 503 “5”s in the list altogether, versus much more than 1011 “2”s in the list. Since the number of “5”s is very limited, there can only be a series of 503 consecutive zeroes at the end of the list.