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Difference between revisions of "1973 AHSME Problems/Problem 19"

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Revision as of 16:19, 5 July 2018

Problem

Define $n_a!$ for $n$ and $a$ positive to be

\[n_a ! = n (n-a)(n-2a)(n-3a)...(n-ka)\]

where $k$ is the greatest integer for which $n>ka$. Then the quotient $72_8!/18_2!$ is equal to

$\textbf{(A)}\ 4^5 \qquad \textbf{(B)}\ 4^6 \qquad \textbf{(C)}\ 4^8 \qquad \textbf{(D)}\ 4^9 \qquad \textbf{(E)}\ 4^{12}$

Solution

Using the definition of $n_a!$, the quotient can be rewritten as \[\frac{72 \cdot 64 \cdot 56 \cdots 8}{18 \cdot 16 \cdot 14 \cdots 2}\] Note that for a given integer $x$, $\tfrac{72-8x}{18-2x} = 4$. Since $0 \le x \le 8$, the quotient simplifies to $\boxed{\textbf{(D)}\ 4^9}$.

See Also

1973 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions
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