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AHSME 1973 Problem 19

    Define na!n_a! for nn and aa positive to be na!=n(n−a)(n−2a)(n−3a)...(n−ka)n_a ! = n (n-a)(n-2a)(n-3a)...(n-ka) where kk is the greatest integer for which n>kan>ka. Then the quotient 728!/182!72_8!/18_2! is equal to (A) 45(B) 46(C) 48(D) 49(E) 412\textbf{(A)}\ 4^5 \textbf{(B)}\ 4^6 \textbf{(C)}\ 4^8 \textbf{(D)}\ 4^9 \textbf{(E)}\ 4^{12}

    Dataset text NuminaMath-1.5. Quoted with source.

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    4^9

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