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

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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>	+
−	~~<math>\textbf{(A) } 10 \ \ \ \ \ \ \textbf{(B) } 13 \ \ \ \ \ \ \textbf{(C) } 16\ \ \ \ \ \ \textbf{(D) } 19\ \ \ \ \ \ \textbf{(E) } 22</math>~~

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