# 1998 AHSME Problems/Problem 30

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## Problem

For each positive integer $n$, let $a_n = \frac{(n+9)!}{(n-1)!}$

Let $k$ denote the smallest positive integer for which the rightmost nonzero digit of $a_k$ is odd. The rightmost nonzero digit of $a_k$ is $\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ }5 \qquad \mathrm{(D) \ } 7 \qquad \mathrm{(E) \ } 9$

## Solution

We have $a_n = n(n+1)\dots (n+9)$.

The value $a_n$ can be written as $2^{x_n} 5^{y_n} r_n$, where $r_n$ is not divisible by 2 and 5. The number of trailing zeroes is $z_n = \min(x_n,y_n)$. The last non-zero digit is the last digit of $2^{x_n-z_n} 5^{y_n-z_n} r_n$.

Clearly, the last non-zero digit is even iff $x_n - z_n > 0$ iff $x_n > y_n$.

Thus we are looking for the smallest $n$ such that the power of $5$ that divides $a_n$ is at least equal to the power of $2$ that divides $a_n$.

The number $a_n$ is a product of $10$ consecutive integers. Out of these, $5$ are divisible by $2$. Out of those $5$, at least $2$ are divisible by $4$, and out of those $2$, one is divisible by $8$. Therefore $x_n\geq 5+2+1=8$ for all $n$.

On the other hand, exactly $2$ of our ten integers are divisible by $5$, and at most one of them can be divisible by a higher power of $5$. As we need $y_n\geq x_n\geq 8$, one of the integers from $n$ to $n+9$ must be divisible by $5^7 = 78125$. Therefore $n\geq 78116$.

We can now take numbers starting with $78116$, and write each of them in the form $2^x 5^y r$. We are looking for 10 consecutive rows where the sum of $y$s is at least equal to the sum of $x$s.

 number   x  y  z
78116    2  0  19529
78117    0  0  78117
78118    1  0  39059
78119:   0  0  78119
78120:   3  1  1953
78121:   0  0  78121
78122:   1  0  39061
78123:   0  0  78123
78124:   2  0  19531
78125:   0  7  1
78126:   1  0  39063


At this point we can stop, as we just found out that $a_{78117}$ is of the form $2^8 5^8 r_{78117}$. Therefore the $k$ we seek is $k=78117$.

Now all we need to do is to compute the last non-zero digit. As the powers of $2$ and $5$ that divide $a_{78117}$ are equal, the last non-zero digit is simply the product of the last digits of the ten $z$s. This is $7\cdot 9\cdot 9\cdot 3\cdot 1\cdot 1\cdot 3\cdot 1\cdot 1\cdot 3 \equiv\boxed{9}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 