2004 AMC 10B Problem 4

P1-10~3 minPrint

Divisibility & Factors GCD & LCM Set up the equation/formula and compute; no special trick needed Consider the largest/smallest element or boundary case

A standard six-sided die is rolled, and $P$ is the product of the five numbers that are visible. What is the largest number that is certain to divide $P$ ?

Show solution without answering

Key insight

Only the bottom face is hidden, so P is 720 divided by that face. Check every possible hidden face and see which factors always survive.

Solution

All six faces multiply to $1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720$ . The five visible faces are all of them except the one on the bottom, so $P$ is $720$ divided by the bottom number.
Make a table of all six cases, and write each product in prime factors so you can compare them:

Bottom face	$P$	Prime factors
$1$	$720$	$2^4 \cdot 3^2 \cdot 5$
$2$	$360$	$2^3 \cdot 3^2 \cdot 5$
$3$	$240$	$2^4 \cdot 3 \cdot 5$
$4$	$180$	$2^2 \cdot 3^2 \cdot 5$
$5$	$144$	$2^4 \cdot 3^2$
$6$	$120$	$2^3 \cdot 3 \cdot 5$

3. "Certain to divide" means it must divide every row, so for each prime take the smallest power that appears in the column.
- The prime $2$ : the fewest $2$ s is $2^2$ (when the $4$ is on the bottom).
- The prime $3$ : the fewest $3$ s is $3^1$ (when the $3$ or the $6$ is on the bottom).
- The prime $5$ : when the $5$ is on the bottom there is no $5$ at all.
4. So the largest number that always divides $P$ is $2^2 \cdot 3 = 12$ .

The answer is $\boxed{\textbf{(B)}\ 12}$ .

Why this works

"Certain to divide" is a worst-case question. For each prime, find the roll that leaves the fewest copies of it, and keep only that many. Writing each case in prime factors makes the comparison easy; this is exactly how you find the greatest common divisor of several numbers.

Alternative approach

Think about which even and which multiple-of-3 faces can be hidden. There are three even faces ( $2, 4, 6$ ); at least two are always visible, and the worst pair is $2$ and $6$ , which give only $2 \times 2 = 2^2$ (hide the $4$ ). There are two multiples of $3$ ( $3$ and $6$ ); at least one is always visible, but hiding one of them can leave just one factor of $3$ . The $5$ can be hidden. So the guaranteed divisor is $4 \times 3 = 12$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The six faces multiply to $1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 720$ . If the hidden bottom face shows $h$ , then the visible product is $P = 720/h$ . As $h$ runs through $1, \dots, 6$ , the possible values of $P$ are
$720,\ 360,\ 240,\ 180,\ 144,\ 120.$

A number is certain to divide $P$ exactly when it divides all six of these, so we want their greatest common divisor. Dividing $720$ by each of $1, \dots, 6$ and taking the gcd is the same as dividing $720$ by the least common multiple of $1, \dots, 6$ :
$\gcd\!\left(\tfrac{720}{1}, \dots, \tfrac{720}{6}\right) = \frac{720}{\operatorname{lcm}(1,2,3,4,5,6)} = \frac{720}{60} = 12.$

Concretely: hiding $5$ kills the factor $5$ ; hiding $4$ leaves only $2^2$ among the powers of $2$ (from $2$ and $6$ ); hiding $3$ or $6$ leaves a single factor of $3$ . What survives every case is $2^2 \cdot 3 = 12$ .

The answer is $\boxed{\textbf{(B)}\ 12}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
Introduction to Number Theory, Chapter 3: Multiples and Divisors (§3.3, §3.4, §3.7)
Introduction to Number Theory, Chapter 5: Divisor Problems (§5.2, §5.3)
Introduction to Number Theory, Chapter 4: Prime Factorization (§4.7)