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

Latest revision as of 13:01, 20 February 2020

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 $\boxed{\textbf{(D)}\ 4^9}$ .

1973 AHSME (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

@@ Line 20: / Line 20: @@
 ==See Also==
-{{AHSME 35p box|year=1973|num-b=18|num-a=20}}
+{{AHSME 30p box|year=1973|num-b=18|num-a=20}}
 [[Category:Introductory Number Theory Problems]]

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

Latest revision as of 13:01, 20 February 2020

Problem

Solution

See Also