Difference between revisions of "2023 AMC 10A Problems/Problem 19"

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@Announcement Ping @offtopic announce ping @random A lot of people have been asking me whether I am joking or not when I say the rumored leak that was posted on math discords and AoPS today was legit. I am not joking, and I have very reasonable evidence to suggest that the leak is legit. The person who has posted this leak has been banned from this server. People have been reporting this individual spamming various math servers and spamming numerous dms. It's pretty clear this individual does not have good intentions. I think this individual is trying to maliciously sabotage the MAA's efforts (wouldn't be the first time something like this has happened).
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Let <math>f(x)</math> be the number of trailing zeroes of <math>x!</math>. Let <math>g(x)=5x+1</math>. Find the sum of the digits of <math>f(1)+f(g(1))+f(g^2(1))+...+f(g^{10}(1))</math> given that <math>5^{11}=48828125</math>.<math>\newline</math>
 
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<math>\textbf{(A) } 10 \ \ \ \ \ \ \textbf{(B) } 13 \ \ \ \ \ \ \textbf{(C) } 16\ \ \ \ \ \ \textbf{(D) } 19\ \ \ \ \ \ \textbf{(E) } 22</math>
Theres been two reputable people who've come to me independently, saying that they asked their proctor, and the proctor confirmed that the test matched. How would the proctor know what the test is? They get access to the test in advance for print and scan administration (see screenshot from MAA's official proctor instructions slideshow).
 
 
 
How would this individual get access to the pdf? I suspect negligence on their proctors part, and somehow they weren't careful with keeping the pdf they received private. I along with other people are trying to contact MAA to resolve this situation. A lot of people still refuse to believe me, and I admit, I first thought this possibility was preposterous. But I have my entire organization at stake with this, and I will take no chances. If you see this individual anywhere, please notify me immediately.
 
 
 
Will the test actually get fixed? I'm worried it might be too late. AMC questions get leaked on stackexchange and wechat all the time, and the MAA hasn't done much to address this. In the end, math competitions are about a learning experience, not about getting a score.
 

Revision as of 16:31, 5 November 2023

Let $f(x)$ be the number of trailing zeroes of $x!$. Let $g(x)=5x+1$. Find the sum of the digits of $f(1)+f(g(1))+f(g^2(1))+...+f(g^{10}(1))$ given that $5^{11}=48828125$.$\newline$ $\textbf{(A) } 10 \ \ \ \ \ \ \textbf{(B) } 13 \ \ \ \ \ \ \textbf{(C) } 16\ \ \ \ \ \ \textbf{(D) } 19\ \ \ \ \ \ \textbf{(E) } 22$