2006 AMC 12B Problem 22

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

Factors of

5

are scarcer than factors of

2,

so

n

counts the factors of

5

in

a!\,b!\,c!

Use

\lfloor x \rfloor + \lfloor y \rfloor

+ \lfloor z \rfloor \ge \lfloor x + y + z \rfloor - 2

at each power of

5

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 \frac{a}{5^k} \rfloor + \lfloor \frac{b}{5^k} \rfloor

+ \lfloor \frac{c}{5^k} \rfloor

\ge \lfloor \frac{2006}{5^k} \rfloor - 2.

Summing over

k = 1, 2, 3, 4

(as

2006 \lt 5^5

) gives

\begin{aligned} &n \ge (401 + 80 + 16 + 3) \\ &\quad {}- 4 \cdot 2 = 492. \end{aligned}

Equality is attainable, for example with

a = b = 624

and

c = 758.

So the minimum is

492.

Thus, the correct answer is B.

Problem 22 in other years