Define n_a! for positive n and a 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

1973 AMC 12 Problem 19

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19.

Define

n_a!

for positive

n

and

a

to be

\begin{aligned} n_a!={}&n(n-a)(n-2a)\\ &\cdot(n-3a)\cdots(n-ka), \end{aligned}

where

k

is the greatest integer for which

n\gt ka.

Then the quotient

\frac{72_8!}{18_2!}

is equal to

4^5

4^6

4^8

4^9

4^{12}

Write both generalized factorials as nine explicit factors

Factor

8

from every numerator term and

2

from every denominator term

The two products are

\begin{aligned} 72_8!&=72\cdot64\cdots8 =8^9(9!),\\ 18_2!&=18\cdot16\cdots2 =2^9(9!). \end{aligned}

Their quotient is therefore

\frac{8^9(9!)}{2^9(9!)}=4^9.

Therefore, the correct answer is D.