Question

Suppose $a,$ $b,$ and $c$ are positive integers with $a + b + c = 2006,$ and $a!\,b!\,c! = m \cdot 10^n,$ where $m$ and $n$ are integers and $m$ is not divisible by $10.$ What is the smallest possible value of $n?$

$489$

$492$

$495$

$498$

$501$

Solution:

Since factors of $2$ are more plentiful than factors of $5,$ $n$ equals the number of factors of $5$ in $a!\,b!\,c!,$ namely $n = \sum_{k \ge 1}\left(\left\lfloor \tfrac{a}{5^k}\right\rfloor + \left\lfloor \tfrac{b}{5^k}\right\rfloor + \left\lfloor \tfrac{c}{5^k}\right\rfloor\right).$

For each $k,$ $\lfloor a/5^k \rfloor + \lfloor b/5^k \rfloor + \lfloor c/5^k \rfloor \ge \lfloor 2006/5^k \rfloor - 2.$ Summing over $k = 1, 2, 3, 4$ (as $2006 \lt 5^5$ ) gives $n \ge (401 + 80 + 16 + 3) - 4 \cdot 2 = 492.$

Equality is attainable, for example with $a = b = 624$ and $c = 758.$ So the minimum is $492.$

Thus, the correct answer is B.

2006 AMC 12B Problem 22

Solution:

Problem 22 in Other Years