Find the greatest possible number of back-to-back zeros at the end of the product of three counting numbers if the sum of these three numbers equals 191. (The number 202100 has exactly 2 back-to-back zeros at the end.)
I tried with examples such as 100,90,1 and 10,81,100, giving me a max amount of 3 trailing zeroes. According to the answer sheet, it said that there were 5 zeroes. Is this possible?
The number of trailing zeros in a number is equal to the largest power of 10 that divides the number (e.g. $202100 = 2021 \times 10^2$). In turn, this is equal to the minimum of the largest power of 2 and 5 respectively that divide the number (e.g. $500 = 2^2 \times 5^3$, $200 = 2^3 \times 5^2$).
Since 2 and 5 are primes, that means we are looking to maximise the total number of factors of each that appear in the three numbers that we will multiply together. For example, the split 100, 90, 1 gives us a total of three 2s and three 5s.
Let's call our numbers $x, y, z$, and we know that $x + y + z = 191$. Since 191 is neither even nor a multiple of 5, at most 2 out of $x, y$ and $z$ can be even or a multiple of 5, but not necessarily the same two. Also, since $191 < 625 = 5^4$, the largest power of 5 that can appear in any of the numbers is 3, and if for example we set $x = 125$ then $y + z = 66$ so then the largest power of 5 we can have in either of them is 2, meaning that our product is definitely capped at having $5^5$ as a factor.
In an attempt to achieve that maximum, we'll leave $x = 125$. Then we want either $y$ or $z$ to have a factor of 25, and also we want $yz$ to have a factor of $2^5$ (because otherwise we won't be able to achieve the required number of factors of 10 in the overall product). Luckily, it's very easy to see that setting $y = 50, z = 16$ achieves the required result.