Solution confusion. Possible overcount?

[tranquilparakeet]

Hi I was confused about the solution to this question.

After accounting for 3, 13, 23, .... 143 once, the solution proceeds to add 23, 30, 73, 130 twice more and adds 123 twice as well.

Why are 23, 30, 73, and 130 counted the same amount as 123(which has 3 factors of 5). Should 23, 30, 73, and 130 be counted once instead of twice?

[debbie]

@tranquilparakeet This is a really good question! It's because when $23$ turns into $25,$ it doesn't just get one more additional factor of $5.$ It gets two more additional factors, since $25$ is not just a multiple of $5$ but is a multiple of $5^2.$ Similarly, $30$ turns into $50,$ so we get another factor of $5^2,$ and the same for turning $73$ into $75.$

Please let me know if this is makes sense or if it's still confusing!

[tranquilparakeet]

@debbie Thanks, however, I am still confused about 123. Since 123 turns into 125 which is 5^3, why does the solution only add three to 125.

Starting, the solution first includes one count for 3, 13, 23 ... 143. Therefore, should the 23(and its multiples), which has already been counted once in the first round, only have one additional number added? For example, in the original problem where we had to find the number of 5's in 147!, we counted all of the multiples of 5's once, then added the multiples of 25 once, and then the multiples of 125 once. Should this same concept be applied to this situation so we add 23, 30, 73, and 130 once and add 123 twice?

[debbie]

@tranquilparakeet Thank you very much! The solution wording was a bit confusing, since it seems to be adding factors over and over again, so I have tried to clean it up a bit by just counting straightforwardly, once, from scratch. (The solution actually did say to count $125$ twice, because it was already counted once when considering the numbers-ending-in-3 that were converted into multiples of 5.)

Please see the list below.

The correct answer given is still correct, $56,$ as there are $21$ additional factors of $5.$

$$\begin{aligned} 3 & \rightarrow 5 && \textcolor{red}{+1} \\ 13 & \rightarrow 15 && \textcolor{red}{+1} \\ 23 & \rightarrow 25 && \textcolor{red}{+2} \\ 30 & \rightarrow 50 && \textcolor{red}{+1} \text{ } \text{ (30 was already a multiple of 5)} \\ 33 & \rightarrow 55 && \textcolor{red}{+1} \\ 43 & \rightarrow 45 && \textcolor{red}{+1} \\ 53 & \rightarrow 55 && \textcolor{red}{+1} \\ 63 & \rightarrow 65 && \textcolor{red}{+1} \\ 73 & \rightarrow 75 && \textcolor{red}{+2} \\ 83 & \rightarrow 85 && \textcolor{red}{+1} \\ 93 & \rightarrow 95 && \textcolor{red}{+1} \\ 103 & \rightarrow 105 && \textcolor{red}{+1} \\ 113 & \rightarrow 115 && \textcolor{red}{+1} \\ 123 & \rightarrow 125 && \textcolor{red}{+3} \\ 130 & \rightarrow 150 && \textcolor{red}{+1} \text{ } \text{ } \text{(130 was already a multiple of 5)}\\ 133 & \rightarrow 155 && \textcolor{red}{+1} \\ 143 & \rightarrow 145 && \textcolor{red}{+1} \\\\ & \text{ Total} && \textcolor{red}{+21} \\ \end{aligned}$$

Please let me know if this makes sense!

[tranquilparakeet]

@debbie ok great. Thank you that makes much more sense!